This problem
Compute $$\sqrt{\sqrt{44\cdot 45 \cdot 46 \cdot 47+1}-44}$$
has a nice solution that relies on the identity
$$n(n+1)(n+2)(n+3) +1 = \left(n^2 + 3n + 1\right)^2$$
a word form of which is
The product of any four consecutive integers is one less than a perfect square.
Is that identity simply a mathematical coincidence - the coefficients just happen to be the same when you expand the left side as when you expand the right - or is there some reason for its truth?
I think this isn't a silly question, because identities like $$(n+1)^2 = n^2 + 2n +1$$ are very intuitive when you draw them:
(the green square represents $n^2$, the light orange rectangles are each $n$, and the dark orange square is $1$ by $1$)
The Key in here is that $$n(n+1)(n+2)(n+3) +1 $$ has a symmetry around $n+1.5$. 1.5 comes as the middle point between $0$ and $3$.
It is
$$(n+1.5-1.5)(n+1.5-0.5)(n+1.5+0.5)(n+1.5+1.5) +1 $$
Now, if we denote $n+1.5=m$ we get
$$(m-1.5)(m-0.5)(m+0.5)(m+1.5)+1$$ or $$(m^2-1.5^2)(m^2-0.5^2)+1=(m^2-2.25)(m^2-0.25)+1$$
Picking $x=m^2-2.25$ this becomes $x^2+2x+1$, which is your picture...