4
$\begingroup$

Find all pair of positive integers $(x,y)$ such that $\frac{x^2+y^2}{x-y}$ is an integer and divides 1995.
Source: 1995 Bulgarian Math Olympiad.

My Work
I think we should start from factorizing $1995 = 3\times 5\times 7\times 19$.
I tried finding some $(x,y)$ which satisfies and then proceed. As when this type of problems appear in Olympiads, most of the time there are only a few solutions. But, I found that $(2,1), (3,1), (21,7), \cdots $ and many many satisfies this. So, I think the answer is a set with some conditions.
I cant find how to start. Any hint will be helpful.

$\endgroup$
4
$\begingroup$

First, we will prove a lemma.

Lemma Let $p$ be a prime number satisfying $p \equiv 3 (mod \ 4)$. Then, $$ p\mid a^2+b^2 \implies p\mid a \ \text{and} \ p\mid b. $$

Proof of Lemma First, note that if $p\mid a$, then $p$ must divide $b$ and vice versa. So, assume $p\nmid a,b$. Now, $$ \frac{a^2}{b^2}\equiv -1 (mod p) $$ gives us a contradiction, since $-1$ is not a quadratic residue modulo $p$, if $p-3$ is divisible by $4$ (well-known fact from number theory).

Equipped with this, we will attack the problem. Suppose, $$ \frac{x^2+y^2}{x-y} = p\alpha $$ for some $\alpha$ and $p \equiv 3 \ (mod \ 4)$. Then, we can define $x = px'$ and $y = py'$ to argue that $$ \frac{x'^2+y'^2}{x'-y'} = \alpha. $$

With this reasoning, starting from any $p\alpha$, we can arrive at either $\alpha = 1$ or $5$, which is the only prime divisor of $1995$, that is congruent to $1$ modulo $4$.

Case 1 $\frac{x'^2 + y'^2}{x'-y'}=1$. With some very basic inequalities, you can convince yourself that this is not possible.

Case 2 $\frac{x'^2 + y'^2}{x'-y'}=5$. In this case, note that $$ x'^2 \leq x'^2 + y'^2 = 5x'-5y' \leq 5x' \implies x'\leq 5. $$

With some algebra, the only solutions are $(x',y') = (3,1)$ and $(2,1)$. Now from here, you can get the entire solution set. For instance, to make $\frac{x^2+y^2}{x-y}=5$, the pairs above work. To make it, say, $3\times 5 \times 7$, multiply these pairs by $3\times 7$. This produces the entire set, and note that you have $16$ possibilities, as $1995$ has 16 divisors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.