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Given a positive integer $k$. Prove that $\forall x;y \in \mathbb{N}; 2k^2-1 \le x <y <2(k+1)^2-1$ then $\frac{x^2+y^2}{2}$ cannot be a perfect square.

My incomplete solution:

From the given conditions, I suppose that it's impossible to solve this by finding the bounds of $\frac{x^2+y^2}{2}$ as $ (2k^2-1)^2 < \frac{x^2+y^2}{2} < (2k^2+4k)^2$.

Therefore, it occurs to me that whether $\frac{x^2+y^2}{2}$ can be a perfect square or not depends on the value of $x;y$. So I tried finding a general formula for $x;y$ so that $\frac{x^2+y^2}{2}$ can be a square of an integer.

Then, I discovered that $x=p(n^2+2mn-m^2);y=p(m^2+2mn-n^2)$ for $p \in \mathbb{N^*};m;n \in \mathbb{N^*}; (m;n)=1; m>n$.

So $2k^2-1 \le p(n^2+2mn-m^2); 2k^2+4k \ge p(m^2+2mn-n^2)$.

But I'm too stupid to continue this solution. So I hope someone would try to solve for me and at your pleasure, I hope you will let me know some books or articles so that I can train on this kind of diophantine equation or some other related kinds of number theory like this.

Bill Dubuque
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khanh20
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1 Answers1

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Assume $x$ and $y$ exist such that $2k^2-1\le x<y<2(k+1)^2-1$ and $\frac{x^2+y^2}2$ is a square number.

First, note that for $\frac{x^2+y^2}2$ to be a square number, it must be an integer, and so $2\mid\left(x^2+y^2\right)$, meaning that $x$ and $y$ are of the same parity. Therefore both $\frac{x+y}2$ and $\frac{y-x}2$ are integers. As mentioned in the link in the OP, $$\left(\frac{x+y}2\right)^2+\left(\frac{y-x}{2}\right)^2=\frac{x^2+y^2}2$$Now, $2k^2-1<\frac{x+y}2<2(k+1)^2-1$, and $\frac{y-x}2<2k+1$. Since the sum of the squares of $\frac{x+y}2$ and $\frac{y-x}2$ is a square, $\frac{y-x}2$ is a difference of two squares, where the lower square is $\left(\frac{x+y}2\right)^2$. This difference is at least $$\left(\frac{x+y}2+1\right)^2-\left(\frac{x+y}2\right)^2=x+y+1>4k^2-1\ge\left(\frac{y-x}2\right)^2$$This means that we must have $\frac{y-x}2=2k$, which clearly has no solutions.

Sai Mehta
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  • Slight note that the final inequality line is not quite true. EG With $ y = 2(k+1)^2 - 3$, $ x = 2k^2 - 1$ (both odd), then $ y - x = 4k$ and so $ (\frac{y-x}{2} )^2 = 4k^2$. But of course, then the $x + y + 1$ is much larger, so you just have to check the bounds more carefully. – Calvin Lin Apr 30 '24 at 18:25