If I know that $b$ is a multiple of $32$ then how can i show that $b+5$ and $b^2 + 7$ are coprime? I'm using the polynomial division method and i got a remainder of 32 but how do I explain in words or mathematically how this justifies my solution.
The previous question has answers based on Properties of Divisibility but I want to know how I can use long polynomial division in my method.
Thanks

Let $d = gcd(b+5, b^{2}+7) = gcd(32k + 5, 32^{2}k^{2} + 320k + 32)$, for some integer $k$. Hence, $d | (b^{2} + 7) - b$, which implies $d | b^{2} + 7 - (b + 5)r$, when $r \in Z.$
Now, we want to choose $r$ such that $(b + 5)r = b^{2} + c$, for some integer $c$; specifically, let $r = (b - 5)$ as then $(b + 5)r$ will result in the difference of two perfect squares $b^{2} - 25$ and permit us to express the difference $b^{2} + 7 - (b - 5)r$ as an integer; that is, $d | b^{2} + 7 - (b^{2} - 25) = 32$.
Certainly $d | b^{2} + 7$ as $b^{2} + 7 = 32^{2}k^{2} + 320k + 32$; however, since $b + 5 = 32k + 5$ and $b = 32k$, it follows that $d | b + 5 - b = 5$; and this allows us to conclude that $gcd(5, 32) = 1.$
Therefore, $gcd(b + 5, b^{2} + 7) = 1$; and so, the two integers contained therein are coprime.
