Basic Number Forms

Question

Show that there exists a perfect square $a^2$ which can also be expressed as the sum of $b^2+4b$ (where $b$ is a positive integer).

Answer 1

If $b$ is a positive integer then $b^2< b^2+4b< b^2+4b+4=(b+2)^2$. The only square between these numbers is $(b+1)^2=b^2+2b+1$.

So if $b^2+4b$ is a square we must have $2b+1=4b\implies b=\frac{1}{2}$. So we conclude $b^2+4b$ is not a square for any positive integer $b$.

Answer 2

If $\;a^2=b^2+4b\;$ , then clearly $\;b<a\;$ and, by parity (either both $\;a,b\;$ are even or both are odd (why?)), $\;b\le a-2\;$ , so

$$a^2=b^2+4b\le(a-2)^2+4(a-2)=a^2-4\;,\;\;\text{which is absurd}$$

Answer 3

If $a^2=b^2+4b$, then $b^2+4b-a^2=0$, so that $b=-2\pm\sqrt{4+a^2}$. In order for $b$ to be an integer, we must have $4+a^2=c^2$ with $c\in\mathbb{Z}$. The only integer squares that differ by $4$ are $a^2=0$ and $c^2=4$, which give $b=-2\pm2\le0$.