Perfect squares in the form $103k-104$ So I am solving a number theory problem and as the last step I have to figure out all perfect squares in the form of $103k-104$ where $k$ is an integer.
I am stuck, is there any hint?
 A: If there is a perfect square of the form $103k-104$, then there is a square $q^2$ satisfying $$q^2 \equiv_{103} -1.$$
So $-1$ would have to be a square in $\mathbb{Z}/103\mathbb{Z}$. But as $103$ is prime, this would imply $103$ must be $1 \pmod 4$, which is not true. [Do you see why this is.]
So there are no such squares.
A: Infinity,
I thought about your problem and this was what I got, I hope it's useful!
if $ k\in\mathbb{Z^+} $ in order to $ 103k-104$ be a perfect square $q$ , at first it needs to positive.
So I defined $\,m = k + 2,\;$where $\;m\in \mathbb{Z^+}$, therefore, it's possible to rewrite your equation as it follows:
$$103(m+2)-104 =q^2 \Leftrightarrow 102+103m=q^2  $$
by doing so, I tried to complete the square
knowing that:$$(a+b)^2=a^2+b^2+2ab$$
if there is a m such that:
$(b+m)^2=q^2$
then we would have:
$$102+103m = (b+m)^2 \implies   \\b^2 = 102,\\ m(2b+m)= 103m$$
and solving this equation, I got:
$m=103-2\sqrt{102}$ which $\in \mathbb{R}-\mathbb{Q}$
But we said earlier that, $m \in \mathbb{Z}$ , so we are in a absurd!
Thank you for the question, I hope I've helped somehow
