Did you know? There is no way to buy exactly 23 items using only nonnegative numbers of five-item and seven-item packets. Yet the equation 5x+7y=23 does have integer solutions—if negative values are permitted. The surprising difference reveals why finding solutions to integer equations requires more than ordinary algebra.
For JC2 students researching an H3 Mathematics tutor in Bukit Timah, this is a useful introduction to linear Diophantine equations and Bézout’s identity. These techniques apply the congruence, proof of existence and construction, and problem-solving principles included in Singapore–Cambridge H3 Mathematics (9820). They are treated here as connected enrichment, not as a separately named H3 additional-content chapter.
Quick answer: When a,b,c are integers and a,b are not both zero, ax+by=c has an integer solution if and only if gcd(a,b) divides c. The extended Euclidean algorithm constructs one solution, and a simple integer parameter describes every solution. Nonnegative or positive constraints introduce another step: checking which members of that family are feasible.
The domain decides what counts as a solution
A linear equation in two unknowns describes a line in the real plane. A Diophantine question only accepts integer lattice points on that line. The line may contain many real points and no integer points, or infinitely many integer points.
If variables count packets, students, machines or objects, they usually need to be nonnegative integers. Negative values might be meaningful for a signed change in stock or a financial balance, but not for the number of packets purchased. Do not silently replace a word-problem domain with all real numbers.
Consider 12x+18y=5. Every integer combination on the left is a multiple of six because both coefficients are multiples of six. The target five is not divisible by six. No integer pair satisfies the equation.
The conclusion is independent of how far one searches. Trying a hundred pairs would still leave infinitely many untested possibilities. A divisibility argument proves all pairs impossible in one step.
Now consider 12x+18y=6. It has a signed solution, such as x=−1,y=1, since −12+18=6. But with x and y both positive, the smallest left side is 12+18=30; it could not equal six. The same equation behaves differently under different permitted domains.
A reliable introduction therefore states the coefficients, the target, the variable domains and any extra bounds before attempting a formula. This may be the most important sentence in an advanced number-theory solution.
Proof of the gcd criterion
Let d=gcd(a,b), the greatest common divisor of integer coefficients a and b, assuming they are not both zero. As d divides both a and b, it divides ax and by for any integers x and y. It therefore divides their sum ax+by.
If ax+by=c has an integer solution, d divides c. This proves necessity. It does not yet prove that every target divisible by d can be formed.
Bézout’s identity supplies the missing direction: there are integers u and v with au+bv=d. The identity follows constructively from the Euclidean algorithm, which reduces the coefficients by successive integer divisions and then reverses those steps.
If d divides c, write c=kd for an integer k. Multiply the Bézout identity by k: a(ku)+b(kv)=kd=c. Thus x=ku and y=kv form an integer solution.
We have proved sufficiency. Together with necessity, this gives the exact if-and-only-if criterion. Students should recognise the proof as two separate obligations rather than treating a checked example as evidence of equivalence.
The special case a=b=0 must be handled separately: the equation becomes 0=c. If c≠0 it is impossible; if c=0, every integer pair solves it. The usual parameter formula is stated for nonzero coefficient pairs.
Worked Euclidean algorithm: coefficients 252 and 198
Start with 252x+198y=18. The Euclidean divisions are 252=198+54, 198=3×54+36, 54=36+18, and 36=2×18+0. Therefore gcd(252,198)=18.
Recover the gcd as a linear combination. The third equation gives 18=54−36. The second gives 36=198−3×54. Substitute to obtain 18=54−(198−3×54)=4×54−198.
Use 54=252−198 to obtain 18=4×252−5×198. Consequently x=4,y=−5 is one solution, and verification gives 1008−990=18.
To solve 252x+198y=90, multiply the identity by five: 90=20×252−25×198. Thus (20,−25) is a seed solution.
Now consider target 100. It is not divisible by 18, so the equation 252x+198y=100 has no integer solutions. One Euclidean computation answers every target for the same coefficient pair.
Back-substitution must be checked carefully. Signs are easy to reverse while changing from a remainder to a coefficient identity. Substitute the final numbers in the original equation before claiming a constructive proof is finished.
A formula for every integer solution
Suppose (x₀,y₀) solves ax+by=c, with nonzero a,b and d=gcd(a,b)>0. If (x,y) is any other integer solution, subtract the two equations to obtain a(x−x₀)+b(y−y₀)=0.
Divide by d, writing A=a/d and B=b/d. These are coprime. The equality A(x−x₀)=−B(y−y₀) means B divides x−x₀, so x−x₀=Bt for an integer t.
Substituting yields y−y₀=−At. Thus all solutions are x=x₀+(b/d)t and y=y₀−(a/d)t, where t may be any integer. This proves both that each parameter choice works and that no solutions are omitted.
For 252x+198y=90, the seed (20,−25) gives x=20+11t, y=−25−14t. At t=−2 the pair (−2,3) satisfies −504+594=90. These examples help verify the formula.
Requiring x,y≥0 produces inequalities: 20+11t≥0 and −25−14t≥0. The first forces t≥−1, while the second forces t≤−2. There is no integer t satisfying both, so this equation has integer solutions but no nonnegative solution.
Change the target to 900. The seed x=y=2 gives 252×2+198×2=900. All solutions are x=2+11t,y=2−14t. The nonnegative restrictions force t=0, showing that (2,2) is the unique nonnegative solution.
The distinction between existence and uniqueness is crucial. A valid seed proves one solution exists; the integer parameter family and its bounds establish whether additional feasible solutions occur.
Use modular arithmetic to solve packet problems
Suppose a shop sells only packets of five and seven items and a customer needs exactly 24 items. Write 5x+7y=24 with x,y≥0 integers. Reduce the equation modulo five: 2y≡4 (mod 5), so y≡2 (mod 5).
Because 7y≤24, a feasible y must be at most three. The only nonnegative number in the residue class two modulo five within that bound is y=2. Substituting gives x=(24−14)/5=2. The only feasible purchase is two packets of each size.
Now change the target to 23. Modulo five, 2y≡3 (mod 5), giving y≡4 (mod 5). The smallest nonnegative possibility y=4 already contributes 28 items, exceeding the target. No feasible purchase exists.
Yet 5(−1)+7(4)=23, so the signed integer equation is solvable. This example makes clear why gcd one establishes signed integer solvability but does not guarantee that a positive target can be formed from nonnegative packets.
Congruences can narrow candidates before enumeration, while coefficient bounds finish a finite feasibility argument. Both are useful H3-style reasoning tools. For an arbitrary large target the parameter interval may contain several integers, not just one.
Related reading: the JC2 H3 modular arithmetic guide explains modular inverses and residues; this article’s distinct purpose is integer solution construction and feasibility.
What a complete JC2 proof should include
First state whether the variables are integers, nonnegative integers or positive integers. Second compute gcd(a,b) and test whether it divides c. Third find and verify a seed when one is possible. Fourth derive the full integer family if completeness is requested.
Fifth translate all real-world constraints into inequalities on the integer parameter. Sixth list or count every allowed parameter. Seventh conclude with the precise property established—existence, impossibility, a unique feasible solution or several solutions.
Watch for a common mistake: if gcd(a,b)=d>1, stepping x by b rather than b/d will generally skip solutions. The coprime reduced coefficients control the smallest valid integer parameter step.
Do not misread the word ‘unique’. For nonzero a,b, a solvable signed integer equation has infinitely many integer solutions because t varies over all integers. A nonnegative restriction may reduce this to zero, one or several points.
Always check the final pair in the original equation. When a proof uses a simplified equation obtained by dividing all coefficients by a common factor, the simplification is valid only when the target is divisible by that factor.
A good H3 tutor asks why a construction covers all possibilities, not only whether a proposed numerical pair is right. That question turns arithmetic into a proof of completeness.
The immutable eduKateSG Mathematics small-group reference documents a three-learner Secondary 1 arrangement near Sixth Avenue MRT; it does not establish a current JC2 H3 timetable. Enquire directly about advanced Mathematics teaching.
Official source: the 2027 H3 Mathematics syllabus 9820 includes proof by construction and congruence. The number-theory examples here are original practice extensions rather than assertions of a separate prescribed chapter.
60 original worked mini-clinics: from gcd checks to complete solutions
1. Gcd obstruction
Question: Can 12x+18y=5 have integer solutions?
Worked answer: No. The left side is always divisible by six, while five is not. This is a universal obstruction, not a failure to search enough pairs.
What it teaches: Use a divisor of both coefficients to rule out every possible integer combination.
2. Another impossible equation
Question: Does 8x+12y=6 have integer solutions?
Worked answer: No. The coefficients’ gcd is four and four does not divide six.
What it teaches: The gcd criterion handles signed integers without trial and error.
3. A signed solution
Question: Solve 12x+18y=6 over integers by giving a seed.
Worked answer: Take x=−1 and y=1: −12+18=6. This is valid even though one coefficient is negative.
What it teaches: A signed solution may be impossible as a physical count but entirely valid in a Diophantine equation.
4. Zero coefficients
Question: What if an equation is 0x+5y=15?
Worked answer: Then y=3, while x is any integer. The zero coefficient leaves x unconstrained.
What it teaches: Handle degenerate coefficients explicitly instead of mechanically applying a formula derived for two nonzero coefficients.
5. Both coefficients zero
Question: Analyse 0x+0y=8.
Worked answer: There is no solution, because the left-hand side is always zero. For target zero, every integer pair would satisfy the equation.
What it teaches: A theorem whose formula divides by gcd needs a separate all-zero case.
6. Euclid first division
Question: What is the first division in gcd(252,198)?
Worked answer: 252=1×198+54. The positive remainder 54 is smaller than 198, so the Euclidean process can continue.
What it teaches: Every remainder equation should be exact; a wrong early quotient corrupts all later coefficients.
7. Euclid second division
Question: Continue the Euclidean algorithm for 252 and 198.
Worked answer: 198=3×54+36. The next division is 54=36+18, followed by 36=2×18.
What it teaches: Successive remainders eventually reveal the gcd.
8. Bézout verification
Question: Check 18=4×252−5×198.
Worked answer: The right side is 1008−990=18, exactly the required gcd.
What it teaches: Always substitute the final coefficients to catch a sign error.
9. Target ninety
Question: Find an integer solution of 252x+198y=90.
Worked answer: Multiplying the gcd identity by five gives x=20 and y=−25. The combination equals ninety.
What it teaches: Bézout sufficiency is constructive: one identity generates target multiples.
10. A second ninety solution
Question: Give another integer pair solving the same target.
Worked answer: Take t=−2 in x=20+11t,y=−25−14t. This gives x=−2,y=3 and −504+594=90.
What it teaches: A parameter family is stronger than a single checked seed.
11. Ninety, no nonnegative pair
Question: Can 252x+198y=90 with x,y≥0?
Worked answer: No. A positive x or y contributes at least 198, while both zero contribute zero. Neither can equal ninety.
What it teaches: A simple lower bound may be more efficient than full parameter inequalities.
12. Nine hundred
Question: Find a nonnegative solution of 252x+198y=900.
Worked answer: x=y=2 works, since 504+396=900. In the complete family x=2+11t,y=2−14t, nonnegativity forces t=0.
What it teaches: The existence and uniqueness parts require different statements.
13. Target one hundred
Question: Can 252x+198y=100 hold over integers?
Worked answer: No. Their gcd is eighteen, which does not divide one hundred.
What it teaches: An exact gcd obstruction is a complete impossibility proof.
14. Two and three target seven
Question: Solve 2x+3y=7 with nonnegative integers.
Worked answer: The only pair is (2,1). The general family x=2+3t,y=1−2t requires t=0 under nonnegativity.
What it teaches: Integer-parameter inequalities establish that no feasible pair was missed.
15. Two and three target eight
Question: Find all nonnegative solutions of 2x+3y=8.
Worked answer: The pairs are (4,0) and (1,2). Checking nonnegative y=0,1,2 produces exactly those solutions.
What it teaches: Finding the first solution is not a proof of uniqueness.
16. Three and five target twenty-one
Question: Solve 3x+5y=21 nonnegatively.
Worked answer: The possibilities are (7,0) and (2,3). Modulo three gives 2y≡0, so y is a multiple of three; the target bounds allow zero or three.
What it teaches: Congruence plus a finite range proves enumeration complete.
17. Two and three target eighteen
Question: List the nonnegative pairs solving 2x+3y=18.
Worked answer: They are (9,0),(6,2),(3,4),(0,6). The permissible y values are the even integers zero through six.
What it teaches: The number of feasible solutions may be more than one.
18. Packets of five and seven
Question: Find a feasible purchase of exactly 24 items.
Worked answer: Two five-item packets and two seven-item packets give 10+14=24.
What it teaches: Verify an integer solution in the original units and domain.
19. Uniqueness of 24 items
Question: Could there be another nonnegative combination of five and seven reaching 24?
Worked answer: Modulo five, 2y≡4 gives y≡2 (mod 5). Since y≤3, only y=2 works, forcing x=2.
What it teaches: A modular restriction plus a bound establishes uniqueness.
20. Target 23 integer solution
Question: Solve 5x+7y=23 with signed integers.
Worked answer: Take x=−1,y=4. The sum −5+28=23.
What it teaches: Gcd one ensures signed solvability; it does not make negative packet counts feasible.
21. Target 23 no packets
Question: Explain why 23 cannot be formed nonnegatively from 5 and 7.
Worked answer: Modulo five, y≡4 (mod 5). The smallest nonnegative possibility y=4 already contributes 28>23.
What it teaches: Use nonnegativity only after finding the necessary residue class.
22. Seven and eleven Bézout
Question: Find integer x,y satisfying 7x+11y=1.
Worked answer: Choose x=−3,y=2, giving −21+22=1.
What it teaches: Coprime coefficients permit a signed representation of one.
23. Thirty-five and twenty-two
Question: Find coefficients for 35x+22y=1.
Worked answer: The pair x=−5,y=8 works since −175+176=1.
What it teaches: Back-substitution may yield relatively large signed coefficients.
24. Nine and six target three
Question: Solve 9x+6y=3 with one seed.
Worked answer: x=1,y=−1 works since 9−6=3. The gcd is three, so solvability is expected.
What it teaches: Constructive solutions complement the gcd criterion.
25. Nine and six target one
Question: Does 9x+6y=1 have integer solutions?
Worked answer: No. Every left-hand value is divisible by three.
What it teaches: A nondivisible target is impossible even if variables are negative.
26. Eighteen and thirty
Question: Give a seed for 18x+30y=12.
Worked answer: x=−1,y=1 gives −18+30=12. All solutions are x=−1+5t,y=1−3t.
What it teaches: The steps five and minus three come from coefficients divided by gcd six.
27. Four and six target two
Question: Does 4x+6y=2 have an integer solution?
Worked answer: Yes: x=−1,y=1 gives −4+6=2. But no nonnegative solution exists because any positive contribution is at least four.
What it teaches: Distinguish an abstract integer claim from a constrained count.
28. Four and six target zero
Question: Solve 4x+6y=0 over integers.
Worked answer: x=3t,y=−2t for any integer t. Substituting gives 12t−12t=0.
What it teaches: A homogeneous equation can have infinitely many signed solutions.
29. Seven and seven
Question: Does 7x+7y=10 have integer solutions?
Worked answer: No. The left side is divisible by seven, while ten is not.
What it teaches: Common factors can be used before any sophisticated manipulation.
30. Target multiple condition
Question: When is 21x+35y=c solvable over integers?
Worked answer: Exactly when seven divides c, because gcd(21,35)=7.
What it teaches: State both directions of an if-and-only-if theorem.
31. Parameter change proof
Question: Why does x=x₀+(b/d)t, y=y₀−(a/d)t always remain a solution?
Worked answer: Its added contribution is a(b/d)t−b(a/d)t=0, so ax+by remains the original c.
What it teaches: Verifying the family is a distinct proof obligation from finding one seed.
32. Parameter completeness
Question: Why does every integer solution belong to the standard family?
Worked answer: Subtract the seed equation and divide by d; coprimality of a/d and b/d forces the difference in x to be a multiple of b/d.
What it teaches: Completeness means no integer lattice points were omitted.
33. Incorrect large step
Question: Why is changing x by b rather than b/d sometimes insufficient?
Worked answer: When d>1, the step b jumps over valid solutions. For 4x+6y=2, x may change by three, not only by six.
What it teaches: A plausible subfamily can be correct yet incomplete.
34. T bounds first case
Question: For x=20+11t, what ensures x≥0?
Worked answer: The inequality t≥−20/11 forces integer t≥−1.
What it teaches: Translate real bounds into integer endpoints using ceilings.
35. T bounds second case
Question: For y=−25−14t, what ensures y≥0?
Worked answer: The inequality t≤−25/14 forces integer t≤−2.
What it teaches: Floor bounds must be combined with other constraints.
36. Empty parameter interval
Question: What if one condition requires t≥−1 and another t≤−2?
Worked answer: No integer t exists, so no feasible pair exists even though signed solutions do.
What it teaches: Contradictory feasibility constraints are mathematically decisive.
37. Unique parameter
Question: For x=2+11t and y=2−14t, which nonnegative integer t survives?
Worked answer: Only zero; the first inequality gives t≥0 and the second gives t≤0.
What it teaches: Counting integer parameter choices proves a feasible solution is unique.
38. Two packets one size
Question: Can 8x+12y=28 with nonnegative integers?
Worked answer: Yes. x=2,y=1 gives sixteen plus twelve, exactly twenty-eight.
What it teaches: A feasible witness should include its unit interpretation.
39. Unreachable packet count
Question: Can 8x+12y=22 over integers?
Worked answer: No. Their gcd is four and four does not divide twenty-two.
What it teaches: A gcd obstruction applies even before nonnegative constraints.
40. Maximum positive restriction
Question: Could 4x+6y=2 with x≥1,y≥1?
Worked answer: No. The left side is at least ten when both variables are positive.
What it teaches: Sometimes monotonicity and a lower bound are the fastest proof.
41. Strict versus nonstrict
Question: Does x=0 qualify when a problem requires a positive integer?
Worked answer: No. Positive means at least one; nonnegative allows zero.
What it teaches: An endpoint condition can change the solution count.
42. Degenerate zero target
Question: Analyse 0x+0y=0.
Worked answer: Every integer pair is a solution because both sides are zero.
What it teaches: Handle zero-coefficient cases explicitly before citing a parameter theorem.
43. Coefficient gcd and divisibility
Question: Why does any d dividing a,b also divide ax+by?
Worked answer: Write a=dp and b=dq. Then ax+by=d(px+qy), an integer multiple of d.
What it teaches: The necessity proof is ordinary factorisation of an arbitrary combination.
44. Sufficiency via Bézout
Question: How does d|c yield a solution?
Worked answer: Write d=au+bv, let c=kd and multiply to obtain c=a(ku)+b(kv).
What it teaches: The proof constructs integers instead of merely asserting they exist.
45. Inverse modulo seven
Question: Solve 3x≡1 (mod 7).
Worked answer: Since 3×5=15≡1, x≡5 (mod 7).
What it teaches: A modular inverse exists when the coefficient is coprime to the modulus.
46. Unsolvable congruence
Question: Solve 6x≡5 (mod 9).
Worked answer: No solution because gcd(6,9)=3 does not divide five.
What it teaches: The gcd solvability criterion also governs one-variable linear congruences.
47. Several residue classes
Question: Solve 6x≡3 (mod 9).
Worked answer: Divide the common gcd three: 2x≡1 (mod 3), so x≡2 (mod 3), namely 2,5,8 modulo nine.
What it teaches: Noncoprime cancellation can reduce the modulus and retain multiple classes.
48. Gcd of signs
Question: Does replacing a coefficient by its negative change the gcd?
Worked answer: No. gcd(a,b)=gcd(|a|,|b|), taken positive when not both coefficients are zero.
What it teaches: The sign of a coefficient changes the equation but not the gcd obstruction.
49. Zero target, nonzero coefficients
Question: Does ax+by=0 always have an integer solution?
Worked answer: Yes: x=y=0 is a seed, and with both nonzero coefficients there is a nontrivial parameter family.
What it teaches: Existence can be trivial while classification remains interesting.
50. Nonnegative solution count
Question: If both coefficients are positive and c positive, can there be infinitely many nonnegative solutions?
Worked answer: No. Each variable is bounded above by c divided by its positive coefficient, so only finitely many integer pairs are possible.
What it teaches: Real-world feasibility makes the infinite signed family finite.
51. Checking x=0
Question: What if 2x+3y=18 and x=0?
Worked answer: Then y=6, a valid nonnegative solution. A strict positivity requirement would exclude it.
What it teaches: Boundary cases often account for one or more solutions.
52. Checking y=0
Question: What if 2x+3y=18 and y=0?
Worked answer: Then x=9, another valid solution. Together with interior points the equation has four nonnegative solutions.
What it teaches: Always check axis intercepts when the problem allows zeros.
53. A fixed target that fails
Question: Can 5x+7y=1 with nonnegative integers?
Worked answer: No, because the least positive contribution is five. Nevertheless gcd(5,7)=1 gives signed integer solutions.
What it teaches: Being coprime does not permit making every small target from positive packets.
54. Comparing c=23 and c=24
Question: Why are these nearby targets different for five- and seven-unit packets?
Worked answer: Twenty-four has feasible (2,2), while twenty-three has no nonnegative solution, despite both signed integer equations being solvable.
What it teaches: The feasible set is discrete and can change abruptly with the target.
55. A proof communication check
Question: How should an H3 solution end after parameterising?
Worked answer: State the exact allowed integer t values and translate them back to x,y with a verification. If none survive, explicitly say no feasible solution exists.
What it teaches: A final mathematical conclusion should match the original domain and question.
56. Fast impossible target
Question: Can 75x+45y=20 over integers?
Worked answer: No, because the coefficients’ gcd is fifteen and fifteen does not divide twenty.
What it teaches: Simplify the structure before attempting long arithmetic.
57. Unique positive pair
Question: Solve 5x+7y=24 requiring x,y>0.
Worked answer: The solution (2,2) works and the residue/bound argument shows no other positive pair does.
What it teaches: Strict positivity matters, but this solution satisfies it as well.
58. A second feasible pair
Question: For 2x+3y=8, give two nonnegative solutions.
Worked answer: (1,2) and (4,0) both work, verifying that a seed does not ensure uniqueness.
What it teaches: Listing independent witnesses is enough to refute a uniqueness claim.
A targeted seven-session tutorial sequence
Session one: identify domains in five packet and lattice problems. Compare integer, nonnegative and positive integer solutions; ask which modelling assumption changes the mathematical answer, and use a gcd obstruction before any trial.
Session two: perform Euclidean division for (252,198), (35,22) and (18,30). Reverse the remainder equations to construct Bézout coefficients, and check each identity in the original integers.
Session three: prove both directions of the gcd criterion with arbitrary coefficients. Make the necessary divisibility direction and the constructive sufficient direction separate, then write a precise if-and-only-if conclusion.
Session four: derive the full integer parameter family by subtracting two solutions. Identify the reduced coefficients after division by the gcd; compare with an incorrect overlarge step that skips solutions.
Session five: turn nonnegative constraints into integer bounds on the parameter. Practise target 90 versus 900 for the same coefficients and explain why only the second has a nonnegative solution.
Session six: use modular arithmetic and upper bounds on packet counts to solve five- and seven-item problems. Differentiate a residue condition that restricts candidates from a construction that establishes feasibility.
Session seven: solve a new equation without naming the method. The student should choose between a gcd obstruction, Euclidean back-substitution and modular enumeration, and then justify completeness of the final set.
Parents can ask what the variables count, whether negative values are permitted, why the gcd matters and how the student knows there are no other answers. The learner should respond in ordinary mathematical language instead of only displaying a calculator output.
The eduKateSG original three-learner Mathematics reference near Sixth Avenue MRT describes a Secondary 1 programme; it is not confirmation of current JC2 H3 classes. Contact eduKateSG for information about advanced Mathematics availability.
Frequently asked questions
Why does gcd decide signed integer solvability?
Every integer combination is divisible by the gcd; Bézout’s identity constructs that gcd as a combination. Multiplying the identity by an integer gives exactly every permissible target multiple.
Is gcd(a,b)=1 enough to buy any number of packets?
No. It ensures all integer targets can be represented using signed coefficients. Nonnegative packet counts impose additional inequalities and can rule out small positive targets.
Why does the solution formula contain a parameter t?
One linear equation in two integers often has an infinite family of lattice points. A parameter records the repeated offset that leaves the left-hand combination unchanged.
What is the extended Euclidean algorithm?
It computes the gcd through successive integer divisions, then reverses the remainder equations to express the gcd as a linear combination of the original coefficients.
How can I find every nonnegative solution?
First find an integer seed and parameterise all integer solutions. Then solve x(t)≥0 and y(t)≥0, keeping only integer values of t satisfying both.
When is a feasible solution unique?
When exactly one integer parameter value satisfies all stated constraints. A seed alone proves existence, not uniqueness.
Are Diophantine equations compulsory named H3 9820 content?
The 2027 outline lists congruence, modular arithmetic and proof principles. This article develops linear Diophantine equations as connected original enrichment, not an additional named syllabus chapter.
How does this help a Question 6 reading passage?
A supplied definition or divisibility theorem can be used to construct and critique solutions. The skills of checking domains, generalising and proving sufficiency transfer to unfamiliar texts.
Should a tutor allow negative answers?
Only when the question’s variables may be negative. Algebraic integer solutions may be valid despite being unsuitable for physical counts or quantities.
Can the gcd be zero?
For the pair a=b=0, the usual Diophantine criterion is handled separately. The equation is either impossible for c≠0 or satisfied by all integer pairs for c=0.
Official reference and related JC2 H3 guides
The 2027 H3 Mathematics (9820) syllabus explicitly includes congruence, modular arithmetic, construction and proof. This article’s Diophantine examples are original extension exercises applying those competencies, not predictions of particular examination questions.
See also existence and uniqueness, congruence and divisibility, and working backwards. The useful habit is to state a condition, construct a witness and establish whether every permissible answer has been captured.
A four-part final proof audit
First check the original domain. An equation in signed integers is not the same as a problem about physical packet counts. Write the restrictions beside the variables before deciding whether a negative solution is valid. An integer solution can be a correct answer to the algebraic problem and a wrong answer to the purchasing problem; both conclusions are consistent because the domains are different.
Next check the exact solvability condition. Compute the greatest common divisor of the two coefficients and determine whether it divides the target. If not, state the obstruction using the fact that every integer linear combination is a multiple of the gcd. If it does, construct at least one solution through Bézout’s identity or an equally valid method. The existence proof is incomplete if it only asserts that a seed must exist.
Then check completeness. Derive the integer parameter family and explain why every other solution differs from the seed by those steps. A list of two or three examples does not establish that no further points occur. When the question imposes nonnegative counts, intersect the parameter family with its integer bounds, including boundary values where one quantity may equal zero.
Finally check interpretation. Report the feasible pairs in the original units, verify each in the initial equation, and distinguish ‘at least one’, ‘exactly one’ and ‘no feasible solution’. The most useful lesson from these problems is not the appearance of a sophisticated number-theory formula. It is the discipline of turning a necessary condition into a sufficient construction, and turning one example into a complete answer.