If $a$ and $b$ are consecutive integers, prove that $a^2 + b^2 + a^2b^2$ is a perfect square. Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the same expression can be written as $$a^2 + (a^2 + 1)(a+1)^2$$ which would imply that there is no real root.  Would really appreciate some help, thanks. 
 A: $$p^2+(p+1)^2+\{p(p+1)\}^2 =p^4+2p^3+3p^2+2p+1 =(p^2+p+1)^2$$

Alternatively,
$$p^2+(p+1)^2+\{p(p+1)\}^2 =(\underbrace{p^2+p})^2+2(\underbrace{p^2+p})+1=?$$
A: $$a^2+b^2+a^2b^2=(a-b)^2+2ab+a^2b^2=1+2ab+a^2b^2=(1+ab)^2$$
A: If $a=b-1$ then square and rearrange to obtain $a^2+2b=b^2+1$. Then
$$a^2+b^2 +a^2b^2=b^2+a^2(b^2+1)=b^2+a^2(a^2+2b)=(a^2+b)^2.$$
A: Since the expression is symmetric in $a$ and $b$, it is not restrictive to assume $b=a+1$, so
$$
a^2+b^2+a^2b^2=
a^2+(a+1)^2+a^2(a+1)^2=
a^4+2a^3+3a^2+2a+1
$$
If you don't see an easy factorization, note that for $a=0$ the statement is clear; for $a\ne0$ we can write
$$
a^4+2a^3+3a^2+2a+1
=
a^2\left(a^2+\frac{1}{a^2}+2a+\frac{2}{a}+3\right)
$$
Now remember that
$$
a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2
$$
so the expression becomes
$$
a^2\left(\left(a+\frac{1}{a}\right)^2+2\left(a+\frac{1}{a}\right)+1\right)=
a^2\left(\left(a+\frac{1}{a}\right)+1\right)^2=(a^2+a+1)^2
$$
A: Let $p=\frac{a+b}{2}$, that is if $a<b$, $p=(2a+1)/2$.
\begin{align}(p-\frac{1}{2})^2+(p+\frac12)^2+(p-\frac12)^2(p+\frac12)^2&=(2p^2+\frac12)+(p^2-\frac14)^2\\
&=p^4+\frac32 p^2+\frac{9}{16}\\
&=(p^2+\frac{3}{4})^2
\end{align}
since $p^2=\frac{4a^2+4a+1}{4}$,
$$p^2+\frac{3}{4}=a^2+a+1.$$
Hence, $$a^2+b^2+a^2b^2=(a^2+a+1)^2$$
A: Note that $a^2+b^2+a^2b^2 = (a^2+1)(b^2+1)-1$
Then this becomes a special case of the Brahmagupta-Fibonacci identity:
$
(x^2+y^2)(z^2+w^2) = (xz+yw)^2+(xw-yz)^2
$
Take $x = a, y = 1, z=b, w = 1$ and you get:
$
(a^2+1)(b^2+1) = (ab+1)^2+(a-b)^2 = (a^2+a+1)^2+1
$
since $b = a+1$.
