# Show that there are infinitely many integral solutions

How do I show that there infinitely man integral solutions of an equation?

For example

$1947*m+264*n=33$

I used the Euclidean algorithm and then back substituted to find one pair of solution $m=3, n=-22$. But I'm not sure how to find more or prove it.

My textbook hints that - these coefficients are not necessarily produced by euclidean algorithm

• you can just add $264$ to $m$ while subtracting $1947$ from $n.$ Then do it again. And again. – Will Jagy Aug 30 '16 at 22:05
• @user1023 Since you found one solution $(m,n) = (3,-22)$, it follows that all (real-valued) solutions are of the form $(m,n) = (3+t, -22 - 1947t/264)$. Can you see a way to choose $t$ so that both $3+t$ and $-22-1947t/264$ are integers? – Erick Wong Aug 30 '16 at 22:12
• @ErickWong I'm sorry - I should have mentioned earlier that m,n are integers – user1023 Aug 30 '16 at 22:14
• @user1023 My comment was already written with the assumption that $m,n$ are integers. Perhaps you didn't understand it? – Erick Wong Aug 30 '16 at 22:15
• @WillJagy thank you - that helps – user1023 Aug 30 '16 at 22:17

## 1 Answer

HINT.-If you have a solution of the equation $ax+by=c$ say $ax_0+by_0=c$ then you have

$$a(x-x_0)+b(y-y_0)=0$$ hence, making $x-x_0=bt$ and $y-y_0=-at$, you have the general solution $$(x,y)=(x_0+bt,\space y_0-at)$$ where $t$ is a parameter.

You have, for instance,$(x,y)=(-13,96)$ (note first that equation reduces to $59x+8y=1$)