Here is my solution . $ a+b-c+d|ab+cd$ $ \Leftrightarrow a+b-c+d|ab+(a+b+d)d=(a+d)(b+d)$ Because $ a+b-c+d>b+d$ so exist a divisor $ k>1$ of $ a+b-c+d$ such that $ k|a+d$t $ \Rightarrow k|b-c$ Because $ n\equiv 1 (\mod 2)$ so $ a^n\equiv -d^n (\mod k )$ $ b^m\equiv c^m (\mod k)$ So $ k|a^nb^m+d^nc^m$ Because $ a^nb^m+c^nd^m>k>1$ so $ a^nb^m+c^nd^m$ is composite .

TTsphn wrote: Here is my solution . $ a + b - c + d|ab + cd$ $ \Leftrightarrow a + b - c + d|ab + (a + b + d)d = (a + d)(b + d)$ Because $ a + b - c + d > b + d$ so exist a divisor $ k > 1$ of $ a + b - c + d$ such that $ k|a + d$t $ \Rightarrow k|b - c$ Because $ n\equiv 1 (\mod 2)$ so $ a^n\equiv - d^n (\mod k )$ $ b^m\equiv c^m (\mod k)$ So $ k|a^nb^m + d^nc^m$ Because $ a^nb^m + c^nd^m > k > 1$ so $ a^nb^m + c^nd^m$ is composite . Why $ a+b-c+d | ab+cd$ ???

Litlle 1000t wrote: Let $ a,b,c,d$ be the positive integers such that $ a > b > c > d$ and $ (a + b - c + d) | (ac + bd)$ . Prove that if $ m$ is arbitrary positive integer , $ n$ is arbitrary odd positive integer, then $ a^n b^m + c^m d^n$ is composite number I think you type problem incorrectly,it should be ''Prove that if $ m$ is arbitrary positive integer , $ n$ is arbitrary odd positive integer, then $ a^n c^m + b^m d^n$ is composite number" please check [Edit:] I've found a solution. same as TTsphn solution from $ 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 \rightarrow a + b = l(a + b - c + d)$ when $ l$ is a divisor of $ k$ if $ l = 1 \rightarrow c = d$ contra! if $ l \geq 2 \rightarrow a + b \geq 2(a + b - c + d) so 2c \geq a + b + 2d$ contra! $ \therefore 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 \rightarrow z | b - c$ $ a \equiv - d (mod z)$ $ b \equiv c (mod z)$ Therefore $ a^n \equiv - d^n (mod z)$ $ b^m \equiv c^m (mod z)$ $ \therefore a^nb^m + c^md^n \equiv 0 (mod z)$ thus it's composite.#