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Can anybody please check my working on this elementary number theory problem?

If there are solutions to $21x+15y=93$ find them.

My work:

Since gcd$(21,15)=3|93$, there are solutions to the Diophantine equation.

gcd$(21,15)=3$, by the extended Euclidean algorithm, we can write $3=-2\times21+3\times15$. Then $31\times3=31\times(-2\times21+3\times15)=-62\times21+93\times15$.

So the only solution is $x=-62$ and $y=93$? Is there any other solutions?

I don't really know any other useful theorems to solve this problem. Can anybody please give some help?

Thanks

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  • $\begingroup$ It becomes $7x + 5y = 31$ $\endgroup$ – N.S.JOHN Mar 30 '16 at 12:56
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Since you know that $21\times (-62)+15\times 93=93$, you can have $$21x+15y=93=21\times (-62)+15\times 93,$$ i.e. $$21(x+62)=15(93-y)$$ Dividing the both sides by $3$ gives $$7(x+62)=5(93-y)$$ Now since $\gcd(7,5)=1$, we have $$x+62=5k,\quad 93-y=7k,$$ i.e. $$x=5k-62,\quad y=-7k+93$$ where $k\in\mathbb Z$.

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  • $\begingroup$ @user71346 : You correctly found one of the solutions. Finding one of the solutions is very important, but that does not mean that there is only one solution. As I wrote, using the solution you found, you can get other solutions. Note that the answer shows that $x=5k-62,y=-7k+93$ are the only solutions (but infinitely many). I hope this helps! $\endgroup$ – mathlove Apr 3 '16 at 7:12
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Every pair $$(5t-62/-7t+93)$$ with $t\in \mathbb Z$ is a solution, so there are infinite many integer solutions.

Additionaly, every integer solution is of the above form for some $t\in \mathbb Z$

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The other solution is $(x,y) = (3,2), (-2,9) \dots$. But observe carefully we can see that there exists a general term: $x=3-5n,y=7n+2$

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