I want to show that $${x^7-1 \over x-1}=y^5-1$$ cannot have any integer solutions. The only observation I have made so far is that the left hand side is the $7$th cyclotomic polynomial $$\Phi_7(x)= {x^7-1 \over x-1}=x^6+x^5+x^4+x^3+x^2+1$$

If I remember correctly cyclotomic polynomials are irreducible. Now can I use this property to arrive at the conclusion or should I try to approach by contradiction and assume $\Phi_7(a)=b^5-1$ for some integers $a$ and $b$? The only problem is that I don't see where I would look for an easy contradiction. Any hints?


I also see that the right hand side can be factored as $$(y-1)(y^4+y^3+y^2+y+1)=(y-1)\Phi_5(y)\implies \frac{\Phi_7(x)}{\Phi_5(y)}=y-1$$

which seems like it could give the result if we prove that the two cyclotomic polynomials have no common factors. How could this be done?

Edit: This is IMO2006 Shortlised Problem N5 (RUS).

  • $\begingroup$ I have no solution to the problem yet, but I noticed that if $y$ is a multiple of $3$, you can prove with modular arithmetic that no integer solutions exist. $\endgroup$ Dec 28, 2015 at 2:10

2 Answers 2


A start to solving this equation is to first prove that if $x$ is an integer and $p$ is a prime divisor of the left hand side $\frac{x^{7}-1}{x-1}$ then either $p=7$ or $p \equiv 1(\mod{7})$.

Proof: First notice by Fermat's little theorem that $x^{p-1}-1$ is divisible by $p$. Also by our hypothesis $x^{7}-1$ is divisible by $p$. Now suppose that $7$ does not divide $p-1$. Then $\gcd(p-1,7) =1$, so there exists integers $a$ and $b$ such that $7a + (p-1)b = 1$. We subsequently see $$ x \equiv x^{7a+(p-1)b} \equiv (x^{7})^{a} \cdot (x^{p-1})^{b} \equiv 1 (\mod p)$$ and so then $$\frac{x^{7}-1}{x-1}= 1 + x + \text{ ... } + x^{6} \equiv 7 (\mod p)$$ So we have that $p$ divides $7$, therefore $p=7$ must hold if we have that $p \equiv 1(\mod 7)$ does not, as we stated. $\square$

We have now shown that every positive divisor of $\frac{x^{7}-1}{x-1}$ meets either $d \equiv 1(\mod 7)$ or $d \equiv 0$.

Now assuming $(x,y)$ is some integer solution to our problem, we notice that $y-1>0$ since $\frac{x^{7}-1}{x-1} > 1 \ \forall \ x \neq 1$. Then since $y-1$ divides $\frac{x^{7}-1}{x-1} = y^{5}-1$ we have $y \equiv 1(\mod 7)$ or $y \equiv 2(\mod 7)$. Evaluating $1+y+y^{2}+y^{3}+y^{4}$ in both possible cases will contradict the fact that our positive divisor $1+y+y^{2}+y^{3}+y^{4}$ of $\frac{x^{7}-1}{x-1}$ is congruent to either $0$ or $1 (\mod 7)$. Your result then follows.

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    $\begingroup$ Can you say more about the first step ? And are you in fact claiming that there is no solution mod 7 ? $\endgroup$ Dec 28, 2015 at 2:29
  • $\begingroup$ @ReneSchipperus I added a proof for my original statement, and I am claiming that this equation has no integer solutions. $\endgroup$
    – miradulo
    Dec 28, 2015 at 2:43
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    $\begingroup$ Quite a tricky proof, with nice insights. $\endgroup$ Dec 28, 2015 at 2:59
  • $\begingroup$ Great answer, thanks! And now I learned that if $\Phi_n(a) \equiv 0 \pmod p$ then $p \equiv 0$ or $1 \pmod n$. $\endgroup$ Dec 28, 2015 at 3:01
  • $\begingroup$ @Sky is that for all $n$ or just prime $n$ ? Do you have a reference ? $\endgroup$ Dec 28, 2015 at 3:03

Since this is an IMO2006 shortlisted problem, here is an solution from The IMO Compendium A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009:

Every prime divisor $p$ of $\frac{x^7-1}{x-1}=x^6+\dots x+1$ is congruent to $0$ or $1$ modulo $7$. Indeed, if $p \mid x-1$, then $\frac{x^7-1}{x-1} \equiv 1+\dots 1\equiv 7 \pmod{p}$, so $p=7$; otherwise the order of $x$ modulo $p$ is $7$ and hence $p\equiv 1 \pmod{7}$. Therefore every positive divisor $d$ of $\frac{x^7-1}{x-1}$ satisfies $d \equiv 0$ or $1\pmod{7}$.

Now suppose $(x,y)$ is a solution of the given equation. Since $y-1$ and $y^4+y^3+y^2+y+1$ divide $\frac{x^7-1}{x-1}=y^5-1$, we have $y\equiv 1$ or $2$ and $y^4+y^3+y^2+y+1\equiv 0$ or $1\pmod{7}$. However, $y\equiv 1$ or $2$ implies that $y^4+y^3+y^2+y+1\equiv 5$ or $3 \pmod{7}$, which is impossible.


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