All integer solutions with a negative coefficient

Goal and scope

All integer solutions of -6x+9y=15 are x=5+3t,y=5+2t, t in Z.

An exact two-variable integer equation, complete parameterization and zero-coefficient cases.

Givens and symbols

• calc

• G1 — a=-6,b=9,c=15; x,y are integers and gcd is nonnegative.

• t — arbitrary integer parameter
• d — nonnegative gcd

Rules and sources

• calc · prerequisite — Assume exact integer arithmetic, division with remainder, and definitions of divisibility and gcd; every congruence has an explicit modulus.
• mko_definition · cited — Use an extended Euclidean Bezout certificate for a particular solution. If d=gcd(a,b) divides c, all solutions are x=x0+(b/d)t and y=y0−(a/d)t, t∈Z. When a=b=0, c=0 permits every integer pair and c≠0 permits none; never divide by d=0. · mko-linear-diophantine-equation

Main derivation

9=1\cdot6+3,\quad6=2\cdot3+0

d=\gcd(-6,9)=3,\quad3=(-6)\cdot1+9\cdot1

-6x+9y=15

3\mid(-6x+9y),\quad3\mid15

3=(-6)1+9(1),\quad15=5\cdot3

x_0=5,\quad y_0=5,\quad-6(5)+9(5)=15

x=5+3t,\quad y=5+2t,\quad t\in\mathbb Z

-6x+9y=15+(-18+18)t=15

(x,y)\text{ any solution},\quad u=x-5,\ v=y-5

-6u+9v=0\iff2u=3v

2u=3v

2u\equiv0\pmod3\Rightarrow u\equiv0\pmod3\Rightarrow u=3t

u=3t,\quad2u=3v

v=2t,\quad x=5+3t,\ y=5+2t

0x+9y=18

y=2,\quad x\in\mathbb Z\text{ arbitrary}

0x+0y=c

c=0\Rightarrow\mathbb Z^2;\qquad c\ne0\Rightarrow\varnothing

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S09 · premise_audit · G1, S08 — For c=0 the solution set cannot be reduced to the one pair (0,0).
• S08 · premise_audit · G1 — Here gcd(0,9)=9 is nonzero, unlike double-zero degeneracy.
• S07 · premise_audit · G1, S04, S05, S06 — S04 proves every parameter is a solution; S05-S07 prove every solution has a parameter.
• S06 · premise_audit · G1, S05 — Cancellation uses the explicitly known inverse 2, not unconditional modular division.
• S05 · inverse_equivalence · G1, S03 — Division by 3 is exact on both sides and loses no solutions.
• S04 · premise_audit · G1, S03 — The negative a makes y's step positive 2, not -2.
• S03 · premise_audit · G1, S01, S02 — d=3 is nonzero and c/d is integral, so scaling is valid.
• S02 · premise_audit · G1, S01 — This is only necessity; the next step constructs a solution.
• S01 · premise_audit · G1 — The original -6 sign is restored, and coefficients 1,1 check directly.

Error controls and negative controls

Completeness

Substitution proves only that the family consists of solutions; subtract a particular solution to prove the converse.

Normalized gcd

d=3, not -3, with step sizes 3 and 2.

• Listing (5,5) gives all solutions. → reject — t=1 gives the distinct solution (8,7).
• The y-step should be -2. → reject — x=5+3t,y=5-2t gives 15-36t, not identically 15.
• Double-zero coefficients allow c/d with d=0. → reject — Zero division is invalid; split on 0=c first.

Result and limitations

All integer solutions of -6x+9y=15 are x=5+3t,y=5+2t, t in Z.

Verification and source binding

2 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: 51507b62feb5851a8cf0c74a7f420b24a166be908dc2a9f02c03852a04af9026
• original source SHA-256: 4e219790cfd274009340791ffd283bde4d21693b21fdc40ee88a5d801a4090e1
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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