Find all integer values of $a$ such that $a^2 - 4a$ is a perfect square

We know that $a^2-4a=(a-2)^2-4$.

It is obvious that these numbers would be nonconsecutive since all consecutive perfect squares differ by odd numbers, so they are nonconsecutive. So, $(a-2)^2=4$ is the only possible outcome. This means that the only solutions are $boxed{0,4}$ as you obtained.

Hope this helped!

If $mge 0$ is such that $m^2 = a^2 - 4a$ we can notice that $m^2 = a^2 - 4a< a^2 -4a + 4 =(a-2)^2 = m^2 + 4$.

So $m < a-2$. Le $(a-2)- m = k$ so that $m+k = a-2$.

so $(m+k)^2 = (a-2)^2$

$m^2 + 2mk + k^2 = a^2 -4a + 4 = m^2 + 4$

$2mk + k^2 = 4$.

but $mge 0$ and $k>0$. There's just not that many choices! $2mk ge 0$ so $k^2 = 4-2mk le 4$ so $kle 2$ so $k = 1, 2$ but $k=1$ means $2m + 1=4$ so $m=frac 32$ is not an integer. So $k =2$ and $m =0$.

That's it $m = 0$ and $m+k = a-2$ so $0+2 = a-2$ so $a=4$.

==== or ....======

If $m^2 = a^2 -4a < a^2$ then $m^2 +4 = a^2 -4a + 4 = (a-2)^2$ is also a perfect square.

Is there ever a case of two perfect squares being exact $4$ apart?

There's something that comes to mind but... I'll ignore it because clever tricks are tricks and if you don't see them right away you should be able to work it out anyway....

But if $m^2 + 4=k^2=(a-2)^2$ then ... two ideas:

1. $k^2 - m^2 = (k-m)(k+m) = 4$ so either $k-m =k+m=2$ and $m=0$ and $k=2$ and $k=a-2$ so $a = 4$.

or $k-m =1$ and $k+m=4$ and $m=frac 32$ and $k=frac 52$ but that's not a perfect integer. Although we do have $(frac 32)^2 = (frac 92)^2 -4cdot frac {9}2$....

2. Le $k=m+i$ and so $m^2 + 4 = k^2 =m^2 +2mi + i^2$ so $2mi+i^2 = 4$ and the only options are $i=2$ and $m=0$ so $k =2$ and $a = k+2 =4$.

.....

Clever trick that I mentioned but didn't want to use...

$n^2 = 1 + 3 + 5 + ...... +(2n-1)$ so if $k^2 - m^2 = 4$ we have a sequence of consecutive odd numbers that add to $4$. And can only be $1+3 = 4$ so $m^2 =0$ and $k^2= 1+3 = 4 = 2^2$ so $m = 0; k=2; a=k+2=4$.

Of course you have to recognize that if $a$ is an integer, $a^2-4a+4$ is already a perfect square, that of $a-2$ and $2-a$.

But the main observation is, if you look at the sequence of squares of whole numbers (including $0$), i.e. $$0^2,1^2,2^2,3^2,4^2,5^2,6^2,7^2cdots \ 0,1,4,9,16,25,36,49,cdots$$ and look at their pairwise differences, i.e. $$1-0,4-1,9-4,16-9,25-16,36-25,49-36,cdots \ = 1,3,5,7,11,13,cdots$$ which gives precisely the odd numbers (since $(n+1)^2-n^2=2n+1$), you can deduce that to get from one square number $S_1 (a^2-4a, in this case)$ to another $S_2 (a^2-4a+4, in this case)$, you always have to make a jump whose magnitude is the sum of a few consecutive odd numbers.

For e.g. to move from $10^2=100$ to $16^2=256$, you have to make the jumps $16^2-15^2=31, 15^2-14^2=29, 14^2-13^2=27,13^2-12^2=25,12^2-11^2=23,11^2-10^2=21$, indeed $16^2-10^2$ can be written as a telescoping sum of the above terms.

P.S. - The actual fun now is sitting down and trying to figure out some obvious patterns, if any ;) for the sequence of cubes, perfect $4^{th}$ powers, perfect $5^{th}$ powers