Show that $4a(a+1)^2$ can be written as $b^4-c^2$, where $a,b,c \in \Bbb Z^+$

Question

Let $a$, $b$ and $c$ be positive integers. Show that any number in the form $4a(a+1)^2$ can be written as $b^4-c^2$ for some $b$ and $c$.

I have shown that any multiple of 4 can be written as the difference of two squares and that any odd integer can be written as the difference of two squares, but can't seem to put them together to prove this. Any help would be appreciated!

Answer 1

Recall that $4xy=(x+y)^2-(x-y)^2$. Now use $x=a(a+1)$ and $y=a+1$. To enforce non negativity of $b$ and $c$, use absolute values: e.g. $(x-y)^2=|x-y|^2$. In fact this gives you positivity of $b$ and $c$ for all $a$ except for when $a=1$.

Answer 2

A few examples with a computer search led to this conjecture, which is easily checked: $$ 4a(a+1)^2=(a+1)^4-(a^2-1)^2$$