# Find all integer solution of the equations:

Find all integer solutions of the equations: \begin{cases} 6x+21y&=&33 \\ 14x-49y&=&13 \end{cases}

I'm not sure how to find all integer solutions for a, but I know there are no integer solutions for b, because $\mbox{gcd}(14,49)=7$ and $7$ does not divide $13$.

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In a), first step is to divide through by 3. – Gerry Myerson Sep 18 '12 at 23:46
I know that.  Which gives 2x+7y=11.  By looking at it I can tell one integer solution is x=2 and y=1, but I don't know how to find all integer solutions. – Pink Panda Sep 18 '12 at 23:47
Good. Now, solve for $x$ and figure out what $y$ has to do to make $x$ an integer. – Gerry Myerson Sep 19 '12 at 0:05

## 1 Answer

For a, can you find values $a$ and $b$ so that $2(x+a)+7(y+b)=2x+7y$? If so, then given one solution (which you have) you can get more by $x'=x+na, y'=y+nb$ for any integer $n$. Then you need to prove that these are all the solutions.

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I'm not sure I follow, the only values for a, and b that would make 2(x+a)+7(y+b)=2x+7y would be a,b=0.  What does this tell me? – Pink Panda Sep 19 '12 at 15:55
@PinkPanda:  that is not true.  You can cancel the terms in $x$ and $y$, leaving $2a+7b=0$.  Doesn't that have a solution in the integers?  In fact, many of them? – Ross Millikan Sep 19 '12 at 16:03
Oh, I see.  It does have many, such as a=-7 and b=2.  Still don't see how I would solve for all integer solutions.  Do I solve for a and then solve for b?  a=-7b/2 and b=-2a/7 – Pink Panda Sep 19 '12 at 16:17
@PinkPanda:  so in addition to your solution of $(2,1)$, you can have $(2-7,1+2), (2-15\cdot 7, 1+15\cdot 2), \ldots$ – Ross Millikan Sep 19 '12 at 16:45
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