Other solutions to the Diophantine equation $21x+15y=93$ besides $x=-62$ and $y=93$?

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

• It becomes $7x + 5y = 31$ – N.S.JOHN Mar 30 '16 at 12:56

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 $k\in\mathbb Z$.
• @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! – mathlove Apr 3 '16 at 7:12
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 $t\in \mathbb Z$
The other solution is $(x,y) = (3,2), (-2,9) \dots$. But observe carefully we can see that there exists a general term: $x=3-5n,y=7n+2$