Are $4ab\pm 1 $ and $(4a^2\pm 1)^2$ coprime? Let $a\ne b$ be two positive integers. Are $4ab+1$ and $(4a^2+1)^2$ coprime always?
Can you find $a$ and $b$ such that they are not coprime?

Edit:
It has been proved that $4ab-1$ is not a divisor of $(4a^2-1)^2$.
Are $4ab-1$ and $(4a^2-1)^2$ always coprime?
 A: Hint $\ $ Counterexamples abound: choose a modulus $\rm\,m\,$ so that $\rm\:mod\ m\!:\ a\equiv b\:$ and $\rm\:4a^2\pm1\equiv 0.\:$ Then $\rm\:4ab\pm1\equiv 4a^2\pm1\equiv 0,\:$ hence $\rm\,m\,|\,4ab\pm1,4a^2\pm1.\:$ 
E.g. for any $\rm\:a,\,$ let $\rm\,m>1\,$ be a divisor of $\rm\,4a^2\pm1,\,$ $\rm\,b = a\!+\!m,\:$ e.g.  $\rm\:a=1,\: m=4\!\pm\!1,\: b = 5\!\pm\!1.$
Remark $\ \ $ Perhaps it will prove a bit instructive to present how I derived the counterexamples. This will yield a precise criterion for coprimality.
$$\rm\begin{eqnarray}\rm (4a^2\!+\!1,\,4ab\!+\!1) &=&\rm (4a^2\!+\!1,\,4a(b\!-\!a)\!-\!(4a^2\!+\!1))\\ &=&\rm (4a^2\!+\!1,\,4a(b\!-\!a))\\ &=&\rm (4a^2\!+\!1,\,b\!-\!a)\ \  via\ \ \  (4a^2\!+\!1,4a) = 1\ \text{ and Euclid's Lemma}\end{eqnarray}$$
This implies the following general criterion
$$\rm (4a^2\!+\!1,(4ab\!+\!1)^2) = 1\ \iff\ (4a^2\!+\!1,\,b\!-\!a) = 1$$
Hence the only counterexamples arise as above: $\rm\ \ 1 < m\,|\,4a^2\!+\!1,\,b\!-\!a$.  
Alternatively, one could employ the follow Brahmagupta sum of squares identity
$$\rm  (1+4a^2)(1+4b^2)\ =\  (1+4ab)^2 + 4\, (a-b)^2 $$
which should lead to a nice viewpoint in terms of Gaussian integer arithmetic.
A: When $a=1$ and $b=6$, $4ab+1=25$ and $(4a^2+1)^2=25$.
When $a=1$ and $b=4$, $4ab-1=15$ and $(4a^2-1)^2=9$.
A: When $b = 4a^3 + 2a$, $4ab+1$ is exactly equal to $(4a^2+1)^2$.  Why would you think that the former could not divide the latter?
On the other hand, it is true that $4ab - 1$ cannot divide $(4a^2+1)^2$.  This is because the latter is expressible as the sum of two relatively prime squares, and no positive integer congruent to $3 \pmod 4$ divides such a number.
A: Let me prove why  $4ab−1$ is not a divisor of $(4a^2−1)^2$ when $a\ne b$. This proof is not mine and for me it is very interesting (it is a famous proof). Let
$$A=\{(x,y) \mid x\ne y,\quad 4xy−1| (4x^2−1)^2\}$$
Fisrtly we show that $A$ is a symmetric relation on $\mathbb{N}$, that is
$$(\forall (x,y)\in A)((y,x)\in A)\quad\quad\quad\quad(1)$$
Let $(x,y)\in A$ be arbitrary. $mod \space 4xy-1$ we have 
$$(4xy)^2 \equiv1$$ and so
$$4y^2-1\equiv4y^2-(4xy)^2=-4y^2(4x^2-1)\equiv0$$
so
$$4xy-1|4y^2-1$$
that is $(y,x)\in A$.
Next we show that:
$$(\forall (x,y)\in A)(\exists (a,b)\in A)(ab<xy)\quad\quad\quad\quad(2)$$
Let $(x,y)\in A$ be arbitrary. Because $A$ is symmetric, without loss of generality we can assume $x<y$. Let:
$$k:=\frac{(4x^2−1)^2}{4xy−1}<\frac{(4x^2−1)^2}{4x^2−1}=4x^2−1$$
$mod\space 4x$ we have
$$ -1 \equiv 4xy-1$$
$$ \to -k \equiv k(4xy-1)=(4x^2−1)^2 \equiv 1$$
$$\to k\equiv-1$$
$$\to k = 4xz -1<4x^2−1$$
we have
$$4xz -1|4x^2−1,\quad z<x<y$$
$$\to (x,z) \in A ,\quad xz<xy$$
So $(2)$ is proved.
Now it is clear that $A$ must be empty.
