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Prove that for any integer $n$, $\gcd (3n^2+5n+7, n^2+1)=1$ or $41$.

The following answer is convoluted because I've intentionally created excess solutions. However, I can't figure out how to eliminate them! Anyone?

Let $d=\gcd (3n^2+5n+7, n^2+1).$ Then $d|[(3n^2+5n+7)-3(n^2+1)]$ $d |(5n+4)$ And $d | [5(3n^2+5n+7)-3n(5n+4)]$ $d |(13n+35)$ And $d |[5(13n+35)-13(5n+4)]$ $d |123$ Therefore, $d= 1$ or $3$ or $41$ or $123$.

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    @robjohn My question isn't actually homework, and if you don't mind to add your proof, I'm sure it would be beneficial. Cheers.2012-10-23

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From your last step, we get that $d = 1,3,41,123$.

Recall that $n^2 \equiv 0,1 \pmod{3} \text{ (Why?)}$ Hence, $3$ (or) $123$ does not divide $n^2+1$.

EDIT

Note that any $n$ is either $0 \pmod{3}$ or $\pm1 \pmod{3}$.

Hence, $n^2 \equiv 0,1 \pmod{3}$. (Recall that if $x \equiv y \pmod{a}$, then $x^k \equiv y^k \pmod{a}$.)

Hence, $n^2 + 1 \equiv 1,2 \pmod{3}$. This means that $3$ does not divide $n^2+1$. Hence, $3$ cannot divide any divisor of $n^2+1$. This enables us to rule out $d=3$ and $d=123$.

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    Oh thank you, now I see it: Because$3$cannot divide $n^2+1$, any multiple of$3$also can't divide $n^2+1$. Thanks Marvis for your response, and I'm glad I created this problem; it's been educational.2012-10-23
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Or, you can write $(-5n+4)(3n^2+5n+7)+(15n+13)(n^2+1)=41 \ .$

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    This is a neat and elegant solution! Now I'm trying to figure out how to apply this intuition to future such problems...2012-10-23
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Hint $\, $ Let $\rm\:d = gcd$, so $\rm\:d\:|\ i^2\!+1,\, 7+5\,i+3\,i^2.\:$ Then, like taking norms of Gaussian integers, $\rm\:mod\ d\!:\,\ i^2\equiv -1\ \Rightarrow\ 0\equiv 7+5\,i+3\,i^2\equiv 4+5\,i\ \Rightarrow\ 0\equiv (4+5\,i)(4-5\,i)\equiv 4^2\!+5^2\equiv 41$

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Suppose that $ (3n^2+5n+7,n^2+1)=(5n+4,n^2+1)\ne1\tag{1} $ then either $ (5n+4,n+i)=(4-5i,n+i)\ne1\tag{2} $ or $ (5n+4,n-i)=(4+5i,n-i)\ne1\tag{3} $ Since $4-5i$ is a Gaussian prime, $(2)\Rightarrow4-5i\,|\,n+i$. That is, $ \frac{n+i}{4-5i}=\frac{(4n-5)+(5n+4)i}{41}\in\mathbb{Z}[i]\tag{4} $ which is true if and only if $n\equiv32\pmod{41}$.

Since $4+5i$ is a Gaussian prime, $(3)\Rightarrow4+5i\,|\,n-i$. That is, $ \frac{n-i}{4+5i}=\frac{(4n-5)-(5n+4)i}{41}\in\mathbb{Z}[i]\tag{5} $ which is true if and only if $n\equiv32\pmod{41}$.

Thus, $(1)$ implies either $ (2)\Rightarrow4-5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{6} $ or $ (3)\Rightarrow4+5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{7} $ Therefore, $ (1)\Rightarrow n\equiv32\pmod{41}\tag{8} $ It is easy to verify that $ n\equiv32\pmod{41}\Rightarrow41\,|\,(3n^2+5n+7,n^2+1)\tag{9} $


Again, $(1)$ implies either $ (2)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\tag{10} $ or $ (3)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\tag{11} $ Therefore, $ (1)\Rightarrow41=(4-5i)(4+5i)\,|\,(3n^2+5n+7,n^2+1)\tag{12} $ Finally, as Pambos points out, the Euclidean Algorithm yields $ (15n+13)(n^2+1)-(5n-4)(3n^2+5n+7)=41\tag{13} $ Therefore, $ (3n^2+5n+7,n^2+1)\,|\,41\tag{14} $