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I want to find the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$

from $n(n-1)(n+1)$ 2 and 3 should divide this expression for all positive n. how can I find the rest? which python says $(2, 3, 6, 7, 9, 42, 14, 18, 21, 63,126)$

memonto
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  • Where do these numbers come from ? – Tom-Tom Nov 16 '15 at 09:21
  • what for $n=6$, $5$ isn't a divisor ? – Nizar Nov 16 '15 at 09:22
  • it should divide for all n>1 – memonto Nov 16 '15 at 09:22
  • I would concentrate on the 126. If you can show that's always a divisor, then all the others follow. So first thing to do is factor 126 into prime powers, and then show each prime power always divides. – Gerry Myerson Nov 16 '15 at 09:24
  • $126=2\times3\times3\times7$. Check in $\mathbb Z_7$ and look the factor $n^2+5$ in $\mathbb Z_3$. – Tom-Tom Nov 16 '15 at 09:26
  • we dont know 126 divides this for all n. it is the probable answer. since it is clear that 6 is a divisor. I guess it is need to show 7 and 9 can divide this $n^7+7n^5+7n^3-15n$ – memonto Nov 16 '15 at 09:28
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    To show $7$ is always a divisor: In $\mathbb{z}_7$ the expression is equivalent to $(n-3)(n-2)(n-1)(n)(n+1)(n+2)(n+3)$ which is always divisible be $7$. – Yiyuan Lee Nov 16 '15 at 09:30
  • thanks @LeeYiyuan . can we say there is one more 3 divisor inside from this (n-3)(n-2)(n-1)(n)(n+1)(n+2)(n+3)? – memonto Nov 16 '15 at 09:57
  • @memonto I doubt so. The factorization allows one to rewrite the expression as $(n - 3)(n - 2)(n -1)(n)(n + 1)(n + 2)(n + 3) + 7k$. While any integer $1 \le m \le 7$ may divide the product of $7$ consecutive integers, it may not necessarily divide $7k$. – Yiyuan Lee Nov 16 '15 at 10:05

1 Answers1

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If $$n\equiv 0 \mod 3,$$ then the factor $$n^2+3\equiv 0 \mod 3.$$ Otherwise $$n^2\equiv1 \mod 3$$ and thus $$n^2+5\equiv 0\mod3.$$ Therefore $$(n^2+3)(n^2+5)\equiv0\mod3.$$ Let us now look modulo $7$ and compute the number for all elements of $\mathbb Z_7$ (which means to check if the number is a multiple of $7$ for $n=0,\dots 6$. You can easily check that it is true. Therefore this number is a multiple of $2\times3\times3\times7=126$ and all its divisors. There are no others since for $n=2$, $n(n^2-1)(n^2+3)(n^2+5)=126\times3$ but for $n=3$, $n(n^2-1)(n^2+3)(n^2+5)=126\times 32$.

Tom-Tom
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