Prove that for all integers $n>1$, $n^4+4$ is composite.
I got $n^4+4=(n^2-2n+2)(n^2+2n+2)$ What should be my next step?
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$n=3 --> 13$ is prime – mathworker21 Jun 13 '18 at 19:41
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$n^4$ or $n^2$? Also please show what you have tried / thought of on the problem. – kingW3 Jun 13 '18 at 19:42
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7If you mean $n^4+4$, just note $n^4+4 = n^4+4n^2+4-4n^2 = (n^2+2)^2-(2n)^2 = (n^2-2n+2)(n^2+2n+2)$ – mathworker21 Jun 13 '18 at 19:42
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3Lookup the Sophie Germain Identity. – dxiv Jun 13 '18 at 19:43
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What means for a number to be composite? (then look back at what you got). – kingW3 Jun 13 '18 at 19:48
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2when is X*Y not composite? – ZKe Jun 13 '18 at 19:56
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1@Zackkenyon when either $X$ or $Y$ is $1$ or fractional. – Rhys Hughes Jun 26 '18 at 23:36
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If you don’t now the factorization of $X^4+4$ as a $\Bbb Z$-polynomial, you’re in good company: Leibniz didn’t, either (or the similar real factorization of $X^4+1$).
This should be taught in high schools: $$ X^4+4=(X^2-2X+2)(X^2+2X+2) $$
Lubin
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It should be taught in high school, but in order to do so you would have to change a lot of other things in the high school math curriculum first. – ZKe Jun 13 '18 at 20:00
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Well, if they can teach $X^3-a^3=(X-a)(X^2+aX+a^2)$, which they do, I see no reason why they shouldn’t teach this lonely but beautiful fact. – Lubin Jun 14 '18 at 00:22
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2@Zackkenyon I've had success teaching this as completing the square yields a difference of squares, which students find memorable. It also generalizes to completing products. – Bill Dubuque Mar 09 '19 at 03:39
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Your inputs are $n\in\Bbb Z, n>1$
Note for $n>1, n\in\Bbb Z, (n^2-2n+2)\ne(n^2+2n+2)$, so there's no way it's a square number,
and that for $n\in \Bbb Z, n^2-2n+2\in\Bbb Z$ and $n^2+2n+2\in \Bbb Z$. So you have two integer factors, both of which are $>1$ for $n>1$, and thus your $n^4+4$ is composite
Rhys Hughes
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Clearly, $n^4+4=(n^2-2n+2)(n^2+2n+2)$. Moreover, $n^2-2n+2=(n-1)^2+1$ and $n^2+2n+2=(n+1)^2+1$; both these integers are greater than $1$ as $n>1$. Thus, $n^4+4$ is a composite number whenever $n>1$.
SHREYAS PAI
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