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I am looking at a fairly simple problem yet I can't figure it out.

Prove that $16x^4+32x^3+32x^2+16x$ is divisible by 96 for every positive integer x.

I've tried factoring, but I cannot figure out how to prove this.

ampersander
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    What did you get when you factored? You merely have to show that $x^4+2x^3+2x^2+x$ is divisible by $2$ and $3$ – J. W. Tanner Oct 17 '21 at 22:22
  • @J. W. Tanner I got this: $16x(x+1)(x^2+x+1)$ – ampersander Oct 17 '21 at 22:27
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    Can you show $x(x+1)$ is divisible by $2$ for every positive integer $x$? $x(x+1)(x^2+x+1)$ is divisible by $3$ for every positive integer $x$? – J. W. Tanner Oct 17 '21 at 22:31
  • And I would need three consecutive numbers in order for it to be divisible by three $(x)(x+1)(x+2)$ – ampersander Oct 17 '21 at 22:31
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    Can you show if $x=3n$ then $3|x$; if $x=3n-1$ then $3|x+1$; and if $x=3n+1$ then $3|x^2+x+1$? – J. W. Tanner Oct 17 '21 at 22:33
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    Ohh, yes, this makes sense. So if x is divisible by 3, then the entire expression is, if x leaves a reminder 2 after division by 3, it is divisible by three due to the fact that x+1 is, and the third case is in the third polynomial. Thank you. – ampersander Oct 17 '21 at 22:43

2 Answers2

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The polynomial factors as $$ 16x(x+1)(x^2+x+1). $$ Since $x(x+1)$ is always even, the polynomial is always divisible by $32$ and it remains to show that it is divisible by $3$. This is easy to check for the cases $x\equiv 0,1,2 \ (\textrm{mod} \ 3).$

subrosar
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$96=32\cdot 3$. Let’s proof $16x^4+16x$ is divisible by $32$. This is the case if and only if $x^4+x$ is divisible by $2$. As taking powers does not change parity this is clear.

Now we still need to show: $16x^4+32x^3+32x^2+16x$ is divisible by $3$. Taking this modulo $3$ yields $x^4+2x^3+2x^2+x$. Note that if $x$ is a multiple of $3$, this trivially is divisible by $3$. If not, then $x^2,x^4$ has rest $1$ and $x,x^3$ have the same rest modulo $3$ (as $1^2=1,2^2=1,1^3=1,2^3=2$). Thus we end up modulo $3$ with $$ 1+2x+2+x = 3+3x$$ is divisible by $3$.

Lazy
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  • Did you mean $x^4+2x^3+\color{red}2x^2+x$? – J. W. Tanner Oct 17 '21 at 22:40
  • Yes, thank you! – Lazy Oct 17 '21 at 22:43
  • Why did taking modulo 3 divide the equation by 16? I have trouble following the proof. If x is divisible by 3, it is obviously divisible by three. But the sentence afterwards - i had problems understanding it. "If not, then x^2, x^4 has rest 1..." what does this mean? – ampersander Oct 17 '21 at 22:54
  • @ampersander: if $x\not\equiv0\pmod3$, then $x^4\equiv x^2\equiv 1$ and $x^3\equiv x\pmod3$ – J. W. Tanner Oct 18 '21 at 02:27
  • @ampersander Modulo 3 we have $16\equiv1$. Also modulo $3$ we have $1^2=1,2^2=1$. So any $x$ that is not divisible by $3$ (so not $0$ mod $3$) satisfies $x^k=x^{k-2}$. – Lazy Oct 18 '21 at 09:48