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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.

Martin Sleziak
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Kay K.
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  • I'd imagine that using a division table would make life much easier. –  May 22 '17 at 01:22

2 Answers2

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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.

Zain Patel
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  • You must've beaten me by seconds! :) – Deepak May 22 '17 at 01:27
  • 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. – CY Aries May 22 '17 at 01:28
  • @CYKwong Thanks, but $a\in\mathbb N$ was given in the condition. – Kay K. May 22 '17 at 01:30
  • 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. – Deepak May 22 '17 at 01:31
  • @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... – Zain Patel May 22 '17 at 01:31
  • Now Deepak's solution is complete – CY Aries May 22 '17 at 01:31
  • One more question - how can we say that $Q(x)$ has all integer coefficients (not rational)? – Kay K. May 22 '17 at 01:36
  • @KayK. I'm not sure what you mean. Using your notation we have $0^{13} + 0 + 90 = 90 = Q(0)a$ and $1^{13} + 1 + 90=2=Q(1)a$. Regardless of what $Q$ has as coefficients we get that $a|90$ and $a|92$. – Zain Patel May 22 '17 at 01:39
  • My question is how can we say $Q(0)$ and $Q(1)$ are integers.. – Kay K. May 22 '17 at 01:40
  • If they aren't then $x^2-x+a$ doesn't divide $x^{13} + x + 90$. As you would have then have just demonstrated a value of $x$ such that your quadratic doesn't divide your other polynomial - in particular their quotient is some non-integer. – Zain Patel May 22 '17 at 01:43
  • In other words, by definition a polynomial, by definition is assumed to mean one with integer coefficients? $(2x + 6)(3x + 1/2) = 6x^2 + 19x + 3$ but we do not say $(2x + 6)|6x^2 + 19x + 3$? Or do we? It's a very important question. ... Of course we'd never question that irrational coefficients aren't allowed... – fleablood May 22 '17 at 02:26
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Set $x = 0$ hence $a \mid 90$

Set $x = 1$ hence $a \mid 92$

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 $a = 2$.

Martin Sleziak
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Deepak
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  • Thanks, but just one more question - how can we say that $Q(x)$ has all integer coefficients (not rational)? – Kay K. May 22 '17 at 01:36
  • @KayK. Why is that relevant? – Deepak May 22 '17 at 01:39
  • 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). – Deepak May 22 '17 at 01:41
  • Okay. Thanks... – Kay K. May 22 '17 at 01:42
  • Still one more question: is the last line true? an odd number can still divide an even number. – Kay K. May 22 '17 at 01:57
  • "The only odd number that can divide an even number is 1" Um, so 7 does not divide 14? – fleablood May 22 '17 at 01:59
  • @fleablood Brainfart. Deleted. Writing new comment – Deepak May 22 '17 at 01:59
  • If $x = -1 $ we have $a+2$ divides $88$ and $4|88$ but $3\not | 88$. – fleablood May 22 '17 at 02:03
  • @fleablood Thanks $x=-1$ gives a neater solution. – Deepak May 22 '17 at 02:06
  • Actually, when we say "a polynomial divides another" does that mean the polynomials have integer coefficients or just rationals? We have (x^2 - x + a)Q(x) = x^13 + x + 90 so if x = -1 we have $(a+2)Q(-1) = 88$ but can Q(-1) have a value of k/3? and does $aQ(1) = 92$ and $aQ(0) = 92$ mean thatn $Q(1)$ and $Q(0)$ must be integers? – fleablood May 22 '17 at 02:20
  • @fleablood I'm beginning to think that we need the condition that the quotient must have integer coefficients (to allow the "smart" solution). Do you agree? – Deepak May 22 '17 at 12:44
  • I think that may be part of the definition "divides" and for many texts it is implied in the word "polynomial" (After all polynomial with rational coefficients q_i = m_i/n_i will have the exact same roots or lack thereof of the poly with integral coeficients m_i.lcm(m_j)/n_i). If we allow rational coefficients then this is the same as solving for x^2 - x +a divides integer poly Kx^13 +Kx + 90K. Which if a ne 2 requires assumptions about K. eg if a ne 2 then 2|K. If a = 1 then 3|K and for all n^2+n + 1 divide K(x^13+x+90) which is no doubt impossible but we can do it in any finite number. – fleablood May 22 '17 at 15:53