Swatting flies with a sledgehammer

Prompted by a recent exchange with Gerry Myerson, I was wondering if anyone has a favorite example of a relatively simple problem with a rather elementary (though perhaps complicated) answer for which there's another answer that relies on an elegant use of a powerful result that's almost certainly beyond the background of the poser of the question.

Solution 1:

If $2^{1/n}$ were a rational $a/b$, with $n>2$, then $a^n=b^n+b^n$, which would contradict Wiles proof of Fermats Last Theorem.

Solution 2:

OK, here's my proof that $49 < 50$.

There's one point in the first quadrant where the circle $x^2+y^2=1$ intersects the line $x=y$, and the tangent line to the circle at that point has slope $-1$ and $x$-intercept $\sqrt{2}$. Every line with the same slope and a slightly smaller $x$-intercept intersects the circle twice, and those with a larger intercept do not intersect the circle.

Now suppose we use $7/5$ as an approximation to $\sqrt{2}$, and draw the line with slope $-1$ and that $x$-intercept. Lo and behold, it passes through the points $(4/5,3/5)$ and $(3/5,4/5)$, which are on the circle. So it's a secant line, not a tangent line. Therefore we conclude that $$ \frac75<\sqrt{2}. $$ Therefore $$ 49 = 7^2 < 5^2\cdot2=50. $$