How to eliminate these extra solutions? (finding the gcd of two expressions)

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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I guess people stopped somewhere. You still have to prove that 41 is indeed, the GCD for some such expressions, without which the solution is incomplete. –  Salahuddin Oct 23 '12 at 7:56
@Salahuddin: Even if $41$ were never a GCD (as it is for $n=32$), the statement would be true. –  robjohn Oct 23 '12 at 8:57
@robjohn Logically, yes, but isn't that like saying that the statement would be true if d had been shown to be 1 or 2 or ... or 100000 ? Do you think it might be a good idea to end such a proof by giving a possible n for each value of d found, to prove that that d is indeed a possible gcd ? –  Ryan Oct 23 '12 at 9:34
@Ryan: point taken. The aim here is to remove all the impossibilities. I retract my statement, except that the GCD is $41$ when $n\equiv32\pmod{41}$. –  robjohn Oct 23 '12 at 12:24
@Ryan: I have a complete proof using Gaussian factorization (which is used in Bill Dubuque's answer) that $(3n^2+5n+7,n^2+1)=41$ iff $n\equiv32\pmod{41}$. I can post if after the homework is due, if you wish. –  robjohn Oct 23 '12 at 16:53

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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Thanks Marvis. Would you mind to elaborate though, as my number theory is very weak. –  Ryan Oct 22 '12 at 19:35
@Ryan: Working $\pmod 3, 0^2=0, 1^2=1, 2^2=1,$ so $n^2+1=1$ or $2$. Therefore $3 \nmid n^2+1$ –  Ross Millikan Oct 22 '12 at 19:39
@Ross Thanks both! I've understood everything except the final bit: how was 123 eliminated? :D –  Ryan Oct 22 '12 at 19:52
@Ryan $3$ divides $123$. So if $123$ were to divide $n^2+1$, then $3$ will divide $n^2+1$, which is what we have ruled out. –  user17762 Oct 22 '12 at 19:56
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. –  Ryan Oct 23 '12 at 7:25

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... –  Ryan Oct 23 '12 at 7:27

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, $$(3n^2+5n+7,n^2+1)\,|\,41\tag{14}$$

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