2004 OIM Problems/Problem 4
Problem
Find all pairs , where
and
are positive integers of two digits each, such that
and
are four-digit perfect squares.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Let and
; then, by subtracting,
, so
. Notice that since
and
are four-digit,
, so
. Thus
.
Clearly , so both
and
are positive; however, as
cannot equal
due to
, we must have
. Since
and
, it is necessary that
. Then
. Solving for
and
results in
Substituting back in:
From the second equation, rearranging yields
Using the Quadratic Formula:
\begin{align*}
a&=\frac{602\pm\sqrt{602^2-4(101^2-4b)}}{2}\\
&=\frac{602\pm2\sqrt{301^2-(101^2-4b)}}{2}\\
&=\frac{602\pm2\sqrt{4b+301^2-101^2}}{2}\\
&=\frac{602\pm2\sqrt{4b+402\cdot200}}{2}\\
&=\frac{602\pm4\sqrt{b+402\cdot50}}{2}\\
&=\frac{602\pm4\sqrt{b+20100}}{2}\\
\end{align*}
Clearly
must be a perfect square, and the only perfect square that allows
to have two digits is
. Thus
, and:
\begin{align*}
a&=\frac{602\pm4\sqrt{b+20100}}{2}\\
&=\frac{602\pm4\cdot142}{2}\\
&=\frac{602\pm568}{2}\\
&=\frac{602-568}{2}\\
&=\frac{34}{2}\\
&=17\\
\end{align*}
Thus the only solution is
, which works (testing yields the two squares
and
).