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Question: $x^{13}+x+90$ is divisible by $x^2-x+a$ $(a\in\mathbb N)$. Find $a$.

What I did was: \begin{align} &x^{13}+x+90 = Q(x)(x^2-x+a)\\ &\text{Let $t$ be one of roots of }x^2-x+a=0\\ &t^2=t-a\\ &t^4=(t-a)^2=t^2-2at+a^2=(1-2a)t+a(a-1)\\ \\ &t^8=((1-2a)t+a(a-1))^2=(1-2a)^2t^2+2(1-2a)a(a-1)t+a^2(1-a)^2\\ &=(1-2a)^2(t-a)+2(1-2a)a(a-1)t+a^2(1-a)^2\\ &=(1-2a)((1-2a)+2a(a-1))t+a^2(1-a)^2-a(1-2a)^2\\ \\ &t^{12}={(1-2a)t+a(a-1)}{(1-2a)((1-2a)+2a(a-1))t+a^2(1-a)^2-a(1-2a)^2}\\ &=(1-2a)^2((1-2a)+2a(a-1))t^2\\ &+ (a(a-1)(1-2a)((1-2a)+2a(a-1))+(1-2a)(a^2(1-a)^2-a(1-2a)^2))t\\ &+ a^2(a-1)(a(1-a)^2-(1-2a)^2)\\ &=(1-2a)^2((1-2a)+2a(a-1))(t-a)\\ &+(a(a-1)(1-2a)((1-2a)+2a(a-1))+(1-2a)(a^2(1-a)^2-a(1-2a)²^2))t\\ &+ a^2(a-1)(a(1-a)^2-(1-2a)^2)\\ &=\left((1-2a)^2((1-2a)+2a(a-1))+a(a-1)(1-2a)((1-2a)+2a(a-1))+(1-2a)(a^2(1-a)^2-a(1-2a)^2)\right)t\\ &+ a^2(a-1)(a(1-a)^2-(1-2a)^2)-a(1-2a)^2((1-2a)+2a(a-1))\\\\ &t^{13}=((1-2a)^2((1-2a)+2a(a-1))+a(a-1)(1-2a)((1-2a)+2a(a-1))+(1-2a)(a^2(1-a)^2-a(1-2a)^2))t^2\\ &+ (a^2(a-1)(a(1-a)^2-(1-2a)^2)-a(1-2a)^2((1-2a)+2a(a-1)))t\\ &=((1-2a)^2((1-2a)+2a(a-1))+a(a-1)(1-2a)((1-2a)+2a(a-1))+(1-2a)(a^2(1-a)^2-a(1-2a)^2))(t-a)\\ &+ (a^2(a-1)(a(1-a)^2-(1-2a)^2)-a(1-2a)^2((1-2a)+2a(a-1)))t\\ &=((1-2a)^2((1-2a)+2a(a-1))+a(a-1)(1-2a)((1-2a)+2a(a-1))+(1-2a)(a^2(1-a)^2-a(1-2a)^2)+a^2(a-1)(a(1-a)^2-(1-2a)^2)-a(1-2a)^2((1-2a)+2a(a-1)))t\\ &-a((1-2a)^2((1-2a)+2a(a-1))+a(a-1)(1-2a)((1-2a)+2a(a-1))+(1-2a)(a^2(1-a)^2-a(1-2a)^2))\\ &=-t-90\\\\ &(1-2a)^2((1-2a)+2a(a-1))+a(a-1)(1-2a)((1-2a)+2a(a-1))\\ &+(1-2a)(a^2(1-a)^2-a(1-2a)^2)\\ &+a^2(a-1)(a(1-a)^2-(1-2a)^2)-a(1-2a)^2((1-2a)+2a(a-1))\\&=-1\\ \\&\text{with an aid of Wlofram Alpha,}\\ &(a-2)(a^5-19a^4+32a^3-20a^2+5a-1)=0\\ &\therefore a=2 \end{align}

Is there a smarter way to get to the answer? Thanks.

(*) Any solution is welcomed, but this question was given to high school students who didn't learn calculus.

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  • $\begingroup$ I'd imagine that using a division table would make life much easier. $\endgroup$ – user384138 May 22 '17 at 1:22
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Set $x = 0$ hence $a | 90$

Set $x = 1$ hence $a | 92$

Find gcd of $90,92$. This is $2$, which means $a$ can be $1$ or $2$

Finally (credit to @fleablood) set $x = -1$. Note that $a+2$ has to divide $88$, so $a = 2$.

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  • $\begingroup$ Thanks, but just one more question - how can we say that $Q(x)$ has all integer coefficients (not rational)? $\endgroup$ – Kay K. May 22 '17 at 1:36
  • $\begingroup$ @KayK. Why is that relevant? $\endgroup$ – Deepak May 22 '17 at 1:39
  • $\begingroup$ Polynomial divisibility between two polynomials with integer coefficients (like in this case) is a very strong condition. If met, it means that no matter what value you set for $x$, the value of one polynomial will be divisible by the value of the other. You don't actually need to consider the coefficients of $Q(x)$ (which is why I never bothered to introduce this in my answer). $\endgroup$ – Deepak May 22 '17 at 1:41
  • $\begingroup$ Okay. Thanks... $\endgroup$ – Kay K. May 22 '17 at 1:42
  • $\begingroup$ Still one more question: is the last line true? an odd number can still divide an even number. $\endgroup$ – Kay K. May 22 '17 at 1:57
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Hint: Set $x=0$ then $a$ must divide $90$. Also setting $x=1$ shows that $a$ must divide $92$; so $a$ must divide the difference $2$.

The reason this works is because $x^2-x+a = x(x-1)+a$ so it is natural to find conditions on $a$ by setting $x=0,1$ so the first term drops out.

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  • $\begingroup$ You must've beaten me by seconds! :) $\endgroup$ – Deepak May 22 '17 at 1:27
  • $\begingroup$ This proves that $a$ is divisible by 2. But we still have to prove that $a=2$ is a solution and $a=-2$ and $\pm1$ are not. $\endgroup$ – CY Aries May 22 '17 at 1:28
  • $\begingroup$ @CYKwong Thanks, but $a\in\mathbb N$ was given in the condition. $\endgroup$ – Kay K. May 22 '17 at 1:30
  • $\begingroup$ I've updated my answer to show that $a=1$ is not a solution (just realised it myself). But negative numbers are excluded by the $a$ is natural condition. $\endgroup$ – Deepak May 22 '17 at 1:31
  • $\begingroup$ @CYKwong - that's why it's a hint! But anyway note that if $a$ is $\leq 0$ then $x^2 - x + a$ has two roots whilst $x^{13} + x + 90$ has everywhere positive derivative. So we must have $a > 0$. Then we need only check between $a=1,2$. Edit: well turns out the question makes this analysis redundant... $\endgroup$ – Zain Patel May 22 '17 at 1:31

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