Algebra contest problem

Suppose $x$ and $y$ are integers. Given $2xy+14y=-53-13x$, what does $xy$ equal? The answer is $-15$, but how do I get that? I feel like I should be able to find this.

• There are many integer solutions to that equation. One of them is $x = -5$, $y = 3$. Are you looking for the smallest (in terms of magnitude) possible value of $xy$? – Brad May 25 '14 at 1:36
• I concur with Brad, another such solution is $-3$. – Display Name May 25 '14 at 1:41
• There is apparently only one possible value for the product xy if x and y are both integers. -5*3=-15. – Yadnarav3 May 25 '14 at 1:41
• @Yadnarav3 What have you tried? – user122283 May 25 '14 at 1:42
• Sorry, my mistake, I took $xy$ to be an integer solution, rather than $x,y\in \mathrm{Z}$ – Display Name May 25 '14 at 1:42

A "general" way of attacking these types of problems is with a method that could be called "completing the rectangle", analogous to completing the square.

We start with $$2xy + 14y = -53 - 13x$$

Re-order the equation

$$2xy + 14y + 13x = -53$$

Attempt to factor the left hand side into a form that looks like $(ay + b)(cx+y)$

$$2xy + 14y + 13x + 91 - 91 = -53$$

$$(2y +13)(x+7) - 91 = -53$$ $$(2y +13)(x+7)= 38$$

Now we can use the prime factorization of $38 = 2\times19$ to attempt to find solutions to this equation.

One way to accomplish this would be to set $2y+13 = 19$ and $x+7 = 2$. This leaves us with $y = 3$, $x = -5$ and $xy = -15$.

However, another way would be to set $2y + 13 = -19$ and $x+7 = -2$. This leaves us with an equally valid solution of $y = -16$ and $x = -9$.

We can continue with $2y+13 = 1$ and $x+7 = 38$ and even $2y+13 = -1$ and $x+7 = -38$. Note that we cannot use $2y + 13 = \pm2$ or $2y + 13 = \pm38$ because the solution to these equations are non-integer $y$.

This technique works well because there are only finitely many possible solutions and they are each easy enough to check.

• Yes! Perfect, thank you. I thought about rewriting it into (2y+13)(x+7), but then didn't where I could go from there. – Yadnarav3 May 25 '14 at 1:46
• So there are four Diophantine solutions. But not all of them give the same product $xy$. – aschepler May 25 '14 at 1:47
• The problem gave five possible choices, only one of them worked. Sorry, I assumed there would only be one possible product because of how the problem was asked. – Yadnarav3 May 25 '14 at 1:55

We have $\displaystyle 2y(x+7)=-(53+13x)\iff x=2y(x+7)+53+12x$ which is odd

Set $\displaystyle x=2z+1\implies y=-\frac{33+13z}{z+4}\iff y+13=\frac{19}{z+4}$

So, the necessary & sufficient condition for $y$ to be an integer is : $z+4$ divides $19$