Find a in N such that 100a + 76 is perfect square

[west coyote]

Let 100a+76 = k^2

$k^2 \equiv 76 \pmod{100}$

[hexed sierra]

k12byda5h

[west coyote]

which is equivalent to
\begin{itemize}
\item $k^2 \equiv 76 \pmod 4 \implies k^2 \equiv 0 \pmod 4$
\item $k^2 \equiv
76 \pmod {25} \implies k^2 \equiv 1 \pmod {25}$
\end{itemize}

[hexed sierra]

k12byda5h

[west coyote]

Firstone, easily check that $2|k$. Second one implies that $k \equiv \pm 1 \pmod{5}$

[hexed sierra]

k12byda5h

[west coyote]

I think you can finish it.

[tulip helm]

@full ruin has given 1 rep to @west coyote

[rose hinge]

When I started I used elementary number theory by Jones& Jones.