Mathematical Reflections 2017, Issue 3 - Problem S414

Problem:
Prove that for any positive integers $a$ and $b$
$$(a^6-1)(b^6-1)+(3a^2b^2+1)(2ab-1)(ab+1)^2$$
is the product of at least four primes, not necessarily distinct.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Let $$N=(a^6-1)(b^6-1)+(3a^2b^2+1)(2ab-1)(ab+1)^2.$$ If $a=b=1$, then $N=16=2\cdot 2 \cdot 2 \cdot 2$. If $b=1$ and $a>1$, then $N=(3a^2+1)(2a-1)(a+1)^2$ is the product of four factors greater than $1$ and so is the product of at least four primes. Assume that $a \geq b>1$. Let $s=a+b$, $p=ab$ and $d=a-b$. Observe that $d^2=(a-b)^2=(a+b)^2-4ab=s^2-4p$. Clearly, $s \geq 4$, $p \geq 4$ and $d \geq 0$. We have
$$\begin{array}{lll} a^6+b^6&=&(a^2+b^2)(a^4+b^4)-(ab)^2(a^2+b^2)\\&=&(a^2+b^2)(a^4+b^4-a^2b^2)\\&=&(s^2-2p)(s^4+p^2-4s^2p). \end{array}$$
Hence, $$\begin{array}{lll} N&=&p^6-(s^2-2p)(s^4+p^2-4s^2p)+1+(3p^2+1)(2p-1)(p+1)^2\\&=&(p^2+4p-s^2)(p^4+2p^3+p^2s^2+p^2-2ps^2+s^4) \\ &=&(p^2-d^2)[(p^2+s^2-p)^2-d^2p^2]\\&=&(p-d)(p+d)(p^2+s^2-p-dp)(p^2+s^2-p+dp). \end{array}$$
Since $a \geq b>1$, then $p-d>1$ and $p+d>1$. Moreover, $p^2+s^2-p-dp=p(p-d)+s^2-p>p+s^2-p=s^2>1$ and $p^2+s^2-p+dp \geq p(p-1)+s^2>1$, so $N$ is the product of four factors greater than $1$ and so is the product of at least four primes.