Statement: For every integer n, if n is greater than or equal to m, then n^5 - 1 is composite. What is the smallest possible integer m > 1 to satisfy this statement?
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HINT: What do you know about factoring $a^n-b^n$? – Brian M. Scott Sep 28 '20 at 20:27
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2Hint: Try factoring the polynomial. – rogerl Sep 28 '20 at 20:27
2 Answers
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It´s easy. you only factorize $n^5-1=(n-1)(n^4+n^3+n^2+n+1)$. In order that both factors be $\geq 2$ you only need that $n\geq3$
Iesus Dave Sanz
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Hypothetically it could be that if $n = 2$ then $n-1 =1$ but it could hypothetically be that $n^4 +n^3 + n +1=2^4 + 2^3 + 2^2 + 2 + 1$ might be composite. It isn't but it could be. In which case the smallest $m$ is $2$. – fleablood Sep 28 '20 at 20:52
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The trick is to realize that $n^5 - 1= (n-1)(n^4 + n^3 + n^2 + n + 1)$. So if $p=n-1$ and $n^4 + n^3 + n^2 + n+1= q$ then $n^5 -1 = pq$. Always.
The thing is $p = n-1$ could be equal to $1$ and $q = n^4 + n^3 + n^2 + n + 1$ *could be equal to $n^5-1$, if $n = 2$. (But we don't really have to worry about $n^4 + n^3+n^2 + n + 1 = 1$ and $n-1 = n^5-1$ if $n$ is positive.)
So as long as $p,q$ are integers (which they are if $n$ is) and, if $n > 0$, then so long as $n-1 \ne 1$ then $n^5 -1 = pq=(n-1)(n^4+n^3 + n^2+ n + 1$ is composite.
The whole "$n^5-1$ is composite for $n \ge m$ for what value of $m>1$" is just a convoluted way of saying "when is $n-1$ not equal to $1$". And $n-1 > 1$ if $n> 2$ means the answer is.... $m =2$.
fleablood
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