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
 A: If $\ b+5\ $ and $\ b^2+7\ $ are not coprime there must be a prime number $p$ with $$b\equiv -5\mod p$$ This gives $$b^2\equiv 25\mod p$$ and because of $$b^2\equiv -7\mod p$$ we have $\ p\mid 32\ $ , which implies $\ p=2\ $. Hence $\ 2\ $ is the only possible common prime factor. But $\ b+5\ $ is odd if $\ b\ $ is a multiple of $\ 32\ $. Hence $\ b+5\ $ and $\ b^2+7\ $ are coprime in this case.
A: $b^2+7 = (b+5)(b-5) + 32$
So any common factor of $b^2+7$ and $b+5$ must also be a factor of $32$.
Since $b$ is a multiple of $32$, both $b+5$ and $b^2+7$ are odd, so common factor is not a multiple of $2$.
Therefore the only common factor of $b+5$ and $b^2+7$ is $1$ i.e. they are coprime.
A: By the Euclidean algorithm $\:\overbrace{(b^{\phantom{|}}\!\!\!+\!5,\,\color{#c00}{b^2\!+\!7})\, =\, (b\!+\!5,\,\color{#c00}{32})}^{\!\!\!\!\!\!\!\!\!\large \bmod b+5:\ \ \ b\ \equiv\ -5\ \ \Rightarrow\ \ \color{#c00}{b^{\Large 2}+7\ \equiv\ 32_{\phantom{|}}}}\, =\, \overbrace{(\underbrace{\color{#0a0}{32n^{\phantom{|}}\!\!\!+\!5}}_{\large b \ =\ 32n},\,32)\, =\, (\color{#0a0}5,32)}^{\large \bmod 32:\ \ \color{#0a0}{32n+5\ \equiv\ 5_{\phantom{|}}}} = 1$
A: Say $d= \gcd$, since $d\mid b^2-25$ we have $$d\mid (b^2+7)-(b^2-25)=32$$
so $d\mid b$. But then $d\mid (b+5)-b=5$ and this can only be if $d\mid \gcd(5,32)=1$.
A: Let $b=32a$ and so $b+5=2^5a+5$ and $b^2+7=2^{10}a^2+7$.
The $1^{st}$ term is a linear polynomial in $a$ with coefficient in $\mathbb{R}$ and its root is in $\mathbb{R}$. But the $2^{nd}$ polynomial does not have any real root. Hence it cannot be factorized as a product of linear polynomial with real coefficient. So they are coprime.
A: Let $d$ be a common factor of $b+5$ and $b^2+7$.  
Then $d$ divides $(b^2+7)-(b+5)(b-5)=32,$ so if $32|b$ then $d|b$.  
But $d|b$ and $d|b+5$ means $d|5,$ and $d|5$ and $d|32$ means $d=1$ since $\gcd(5,32)=1$.
A: 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.
