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Prove that $$x+y = (x-y)^2$$ has infinite integer solutions.

I tried to reform the equation in several ways. As $$(x-y)(x-y-1)=2y$$ Or $$(x+y)(x+y-1)=4xy$$

I was trying to find y in terms of x But as I saw it wasn't that easy.

Please guide me a way to prove this.

Delta Account
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5 Answers5

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if $x,y$ are integers, then we must have $x+y$ is a perfect square since $x+y=(x-y)^2$. We can then set: $$\begin{cases} x+y =k^2 \\ x-y = \pm k \end{cases}$$ Then solving gives: $$\begin{cases} x=\frac{k^2 \pm k}{2} \\ y=\frac{k^2 \mp k}{2} \end{cases}$$ Its easy to prove with modular arithmetic or simply by going case by case writing $k=2n$ and $k=2n+1$ that $k^2 \pm k$ is divisible by $2$ so $x$ and $y$ are integers as required and there exists solutions for each $k$. You might want to try find $x$ and $y$ for different k to convince yourself this really works.

Ogglie Ostrich
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Let $A=x-y$. Thus $x+y=A+2y$ and the equation becomes equivalent to $$ A^2-A-2y=0 $$ which has solutions $A=\frac12(1\pm\sqrt{1+8y})$.

Thus every time $1+8y$ is a square you get integer solutions.

Andrea Mori
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For each even $n$, write $((n+k)-(0+k))^2 = n^2,$ and since $n^2$ is even also, there must exist an integer $k$ such that $n+2k = n^2.$

Adam Rubinson
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Let $n = x - y$ so that $x + y = (x - y) + 2y = n + 2y$. Substituting, we obtain: $$ n + 2y = n^2 $$ Solving for $y$, we have: $$ y = \frac{n^2 - n}{2} = \frac{n(n - 1)}{2} $$ Solving for $x$, we have: $$ x = y + n = \frac{n^2 - n}{2} + \frac{2n}{2} = \frac{n^2 + n}{2} = \frac{n (n + 1)}{2} $$ Notice that, for both $x$ and $y$, the numerator is a product of consecutive integers if $n \in \mathbb Z$, so one of them must be divisible by $2$. So we've found an infinite family of integer solutions, as desired.

Adriano
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Let $(x-y)=k$. Your equation is then finding $x+y=k^2$, and we know that $x=y+k$. So $k^2-k-2y =0$. Solving for $k$, you get that $k = 1/2\pm1/2\sqrt{1+8y}$. Thus, every time $1+8y$ is a perfect square, you have an integral solution. This is true for every perfect square (hint: write out the expression for $(2x+1)^2$ and you should see it), so your equation has an infinite number of solutions.

Boshu
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