Given $a,b,c,d\in\mathbb{Z}_{+}$ with $a\geqq b>c>d\geqq 0$ and $ac+bd=(\!b+d+a-c\!)(\!b+d-a+c\!)$. Prove $ab+cd$ is not prime. Given positve integers $a, b, c, d$. For $a\geqq\!b>\!c>\!d\geqq\!0$ and $ac+\!bd= (\!b+ d+ a- c\!)(\!b+ d- a+ c\!)$
Prove $ab+ cd$ is not a prime number.
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We have
$$(\!ab+ cd\!)(\!ab- bc- cd\!)\!= (\!b^{2}- c^{2}\!)(\!c^{2}- ca+ a^{2}\!)+ c^{2}\left (\!ac+ bd- (\!b+ d- a+ c\!)(\!b+ d+ a- c\!)\!\right )\!=$$
$$= (b^{2}- c^{2})(c^{2}- ca+ a^{2})$$
where  $(\!2a+ b- c+ 2d\!)(ab- bc- cd)= (\!b- c\!)\left (\!a^{2}+ b^{2}+ c^{2}+ d^{2}+ bc+ 2ad+ (\!b- c\!)(\!a+ d\!)\!\right )+$
$$+ (b+ c)\left (ac+ bd- (b+ d- a+ c)(b+ d+ a- c) \right )=$$
$$= (b- c)\left (a^{2}+ b^{2}+ c^{2}+ d^{2}+ bc+ 2ad+ (b- c)(a+ d) \right )> 0$$
$$\therefore\,ab- bc- cd> 0$$
$$(a+ b- c+ d)\left ( (b^{2}- c^{2})- (ab- bc- cd) \right )= (b- c)(c- d)(d+ a- c)-$$
$$- b\left ( ac+ bd- (b+ d- a+ c)(b+ d+ a- c) \right )=$$
$$= (b- c)(c- d)(d+ a- c)> 0\,\therefore\,b^{2}- c^{2}> ab- bc- cd$$
$$(c^{2}- ca+ a^{2})- (ab- bc- cd)= cd+ c^{2}- c(a- b)+ (a- b)^{2}+ b(a- b)> 0$$
$$\therefore\,c^{2}- ca+ a^{2}> ab- bc- cd$$
q.e.d

We prove that $a^nb^m+c^md^n$ is composite for odd $n$ and arbitrary $m.$
$ a + b - c + d | ac + bd \rightarrow a + b - c + d |(a + b)(a + d)$
$ \therefore \exists k \in N$ ; $ (a + b)(a + d) = k(a + b - c + d)$
Assume that $ gcd(a + d,a + b - c + d) = 1 \implies a + b = l(a + b - c + d)$, where $l$ is a divisor of $ k.$
If $ l = 1\implies c = d$, a contradiction.
If $ l \geq 2 \rightarrow a + b \geq 2(a + b - c + d)$, so $2c \geq a + b + 2d$ which is again a contradiction.
Hence, $ \gcd(a + d,a + b - c + d) > 1.$
Let $ \gcd(a + d,a + b - c + d) = z > 1$
so $ z|a + d ,z|a + b - c + d \implies z | b - c.$
Thus $ a \equiv - d (\mod z)$ and $ b \equiv c (\mod z).$
Finally \begin{cases} a^n \equiv - d^n (\mod z)\\
 b^m \equiv c^m (\mod z)
\end{cases}
Therefore, $ a^nb^m + c^md^n \equiv 0 (\mod z)$ 
