Lemma. If $p, q$ are primes and $q|\frac{x^{p}-1}{x-1}$ then $q \equiv 1 \mod{p}$ or $q \equiv 0 \mod{p}$. Proof: well-known, it was posted many times also. Suppose that $x \not \equiv 1 \mod{7}$. Then of course $\frac{x^{7}-1}{x-1}\equiv \frac{x-1}{x-1}= 1 \mod{7}$. So $y^{5}-1 \equiv 1 \mod{7}$, it implies that $y \equiv 4 \mod{7}$, as one can easily check. But then: $y^{5}-1 = (y-1)\frac{y^{5}-1}{y-1}$ and $y-1 \equiv 3 \mod{7}$. It is contradiction with the lemma, since $y-1$ must have a prime divisor $q$ which is not congruent to $1$ or $0 \mod{7}$. Suppose that $x \equiv 1 \mod{7}$. Then $\frac{x^{7}-1}{x-1}= x^{6}+...+1 \equiv 0 \mod{7}$. Then, we must have $y \equiv 1 \mod{7}$. But: $y^{5}-1=(y-1)(y^{4}+...+1)$ and $y^{4}+...+1 \equiv 5 \mod{7}$. One more time, a contradiction with the lemma.

TomciO wrote: It is contradiction with the lemma, since $y-1$ must have a prime divisor $q$ which is not congruent to $1$ or $0 \mod{7}$. . I do not see the reason.

It's clear that $y-1$ is not divisible by $7$. So, suppose that all prime divisors of $y-1$ are $1 \mod{7}$. Then of course $y-1 \equiv 1 \mod{7}$ as well, but we know that $y-1 \equiv 3 \mod{7}$.

TomciO wrote: So, suppose that all prime divisors of $y-1$ are $1 \mod{7}$. But how does this follow from the lemma?

This follows from the lemma for $p=7$ and the fact that $y-1|\frac{x^{7}-1}{x-1}$ (and of course $7 \not | y-1)$.

Fact (posted often enough before): $\frac{x^{7}-1}{x-1}$ has only prime divisors $\equiv 0$ or$1\mod 7$. Then you want $\frac{x^{7}-1}{x-1}= (y-1)(y^{4}+y^{3}+y^{2}+y+1)$, thus $y \equiv 1,2 \mod 7$ by the fact. Then $y^{4}+y^{3}+y^{2}+y+1 \equiv 5,3 \mod 7$, contradicting the fact. [Note: this post was merged from another topic]

e.lopes wrote: (new topic, because our dear moderator ZetaX have ZERO TOLERANCE! ) Prove that the equation $\frac{x^{7}-1}{x-1}= y^{5}-1$ don't have integer solutions! This problem also from Russian TST 2007(author - A. Efimov). And for every prime $p=3k+2$ the equation $\frac{x^{7}-1}{x-1}=y^{p}-1$ hasn't solutions in $\mathbb{N}$. Funny, first we wanted to give this problem for $p=11$ but I find that this equation hasn't solutions modulo 23!

TomciO wrote: Lemma. If $p, q$ are primes and $q|\frac{x^{p}-1}{x-1}$ then $q \equiv 1 \mod{p}$ or $q \equiv 0 \mod{p}$. Proof: well-known, it was posted many times also. When you say "posted many times", can you give a link then please? because new people like me do not know so many posts and i can not find this when i use search function this is not only here, it happend quite often to me when i looked at this forum in past some days since i found this forum, its a bit frustrating.

For example: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1939040862&t=114452

i couldn't see why the lemma is true ???

i didn't know the lemma and i tried modulo 11... the problem would hold if $y \neq 0$ (mod 11) or $x\neq 7,9$ (mod 11)... i would appretiate if somebody could be so gently to try to finish this up...

Thank you to you Tomic0. campos, it will not work with modulo or they would have taken a different prime... read this: pavel kozlov wrote: Funny, first we wanted to give this problem for $p=11$ but I find that this equation hasn't solutions modulo 23!

TomciO wrote: Lemma. If $p, q$ are primes and $q|\frac{x^{p}-1}{x-1}$ then $q \equiv 1 \mod{p}$ or $q \equiv 0 \mod{p}$. Proof: well-known, it was posted many times also. hi, if p,q are primes and $q \equiv 0 \bmod{p}$ then $p=q$.

Yes, what's wrong about that?

Or we have also at http://www.mathlinks.ro/viewtopic.php?t=123665&search_id=1784689764&start=20 an essentially quite similar problem

isn't this problem trivial if you consider the cyclotomic fields of $ x^7 - 1$ and $ y^5 - 1$????????

Doesn't it follow right from the axioms of arithmetic¿ A more concise and explaining answer would be helpful.

The lemma is very interesting. But I don't prove it . Can you help me

I will use the same formulation as I have used before. So, supose that $ q| \frac{x^{p}-1}{x-1}$. Let $ k$ be the order of $ x$ modulo $ q$. If $ k=1$ then $ x \equiv 1 \pmod{q}$ and $ \frac{x^p-1}{x-1} \equiv x^{p-1} + \ldots + 1 \equiv 1 + 1 + \ldots + 1 \equiv p \equiv 0 \pmod{q}$, so $ p=q$. If, $ k>1$ we have that $ x^p \equiv 1 \pmod{q}$, so $ k|p$ and it means that $ p=k$ since $ p$ is prime. So $ p|q-1$ from the Fermat's Little Theorem.

XOINK. Suppose $p$ is a prime dividing the left side. Then $x^7 \equiv 1 \pmod p.$ Then $x$ has order either $1$ or $7 \pmod p.$ If it has order $1$ then by LTE we get $\nu_p(x^7-1)=\nu_p(x-1)+\nu_p(7)$ so $\nu_p(7)\ge 1$ so $p=7.$ Otherwise $x$ has order $7$ implying that $7 \mid \varphi(p)=p-1,$ so $p \equiv 1 \pmod 7.$ Now factor the right side as $(y-1)(y^4+y^3+y^2+y+1).$ Now $y-1 \equiv 0,1 \pmod 7$ so $y \equiv 1,2 \pmod 7$ and we get $y^4+y^3+y^2+y+1 \equiv 5,3 \pmod 7,$ but it must also be $0,1 \pmod 7,$ contradiction.

We first present a brief lemma on cyclotomic polynomials, following from v_enhance's "Orders Modulo A Prime". Lemma: For any prime dividing $\Phi_7(a)$ we either have $p \equiv 1 \pmod{7}$ or $p \mid 7$. Proof. Take a prime $p$ dividing $\Phi_7(a)$. Note that clearly we must have $$a^7 \equiv 1 \pmod{p}$$as $\Phi_7(X) \mid X^7 - 1$. However then $\text{ord}_p(a) \mid 7$. If $\text{ord}_p(a) = 1$ then we have $a \equiv 1\pmod{7}$, and hence $$\Phi_7(a) \equiv 7 \equiv 0 \pmod{p}$$hence $p = 7$. Otherwise we have $\text{ord}_p(a) = 7$ and so $7 \mid p - 1$, which rearranges to $p \equiv 1 \pmod{7}$ as desired. $\blacksquare$ Now armed with the lemma we will show that this equation has no integer solutions. Note that $\Phi_7(x) = \frac{x^7-1}{x-1}$. Then from the lemma for any prime $p$ dividing $\Phi_7(x)$ for some $x$, we have one of the following: Either $x$ has order $7$ modulo $p$ and hence $p \equiv 1 \pmod{7}$ Or we must have $p = 7$ Now assume $7 \mid \Phi_7(x)$. Then we require $y \equiv 1 \pmod{7}$. Then clearly $\nu_7(y^5 - 1) = 1$ from LTE, and thus all other primes dividing $\Phi_7(x)$ satisfy the first condition. Then must have, $\frac{y^5-1}{y-1}$ congruent to $1$ modulo $7$, as it is only divisible by primes congruent to $1$ modulo $7$. However for $y \equiv 1 \pmod{7}$ this fails. Then assume that $7 \nmid \Phi_7(x)$. From similar reasoning we also must have $y^5 - 1$ be congruent to $1$ modulo $7$ as it is only divisible by primes that are congruent to $1$ modulo $7$. Then we require $y \equiv 4 \pmod{7}$. However factoring the right hand side we can see the factor $(y-1)$ is then not congruent to $1$ modulo $7$ and hence there exists some prime that is not congruent to $1$ modulo $7$ dividing the left hand side, contradiction. Thus there are: $\boxed{\text{No solutions}}$. Remark: I need to go over the general derivation of the lemma for $\Phi_n(X)$, rather than the specific case of $n$ prime, but it seems like it follows a similar idea.

It is well-known that if $\Phi_n(a) \equiv 0 \pmod p$, then either $n \equiv 0 \pmod p$ or $p \equiv 1 \pmod n$. Let $p$ be a prime dividing $\frac{x^7-1}{x-1} = \Phi_7(x)$. Then either $p = 7$ or $p \equiv 1 \pmod 7$, so all factors of the LHS are either $0$ or $1$ $\pmod 7$. $y^5-1$ can be factored as $(y-1)(y^4+y^3+y^2+y+1)$, so we must have $y \equiv 1$ or $2 \pmod 7$. If $y \equiv 1 \pmod 7$, then $y^4 + y^3 + y^2 + y + 1 \equiv 5 \pmod 7$, contradiction. If $y \equiv 2 \pmod 7$, then $y^4 + y^3 + y^2 + y + 1 \equiv 3 \pmod 7$, also contradiction. Therefore, there are no solutions.

We first gonna prove the following "well-known" lemma: $$p\mid \Phi_n(a)\Rightarrow\begin{cases} \operatorname{ord}_p(a)=n \rightarrow n\mid p-1 \hspace{0.3cm}\text{or} \\ p\mid n & \end{cases}$$as $a^n-1=\prod_{d\mid n} \Phi_d(a)$ then $r=\operatorname{ord}_p(a)\mid n$, we got two cases $r=n\Rightarrow n\mid \varphi(p)=p-1$ $r<n \Rightarrow \exists \hspace{0.1cm} d\mid r \hspace{0.2cm} \text{such that}\hspace{0.2cm} p\mid \Phi_d(a)$ because $$0\equiv a^r-1\equiv \prod_{d\mid r} \Phi_d(a)$$Thus $a$ is double root of $x^n-1 =\prod_{d\mid n} \Phi_d(x) \pmod p$ so $a$ is a root of it´s derivative, this gives that $n\cdot a^{n-1}\equiv 0\pmod p$ so $p\mid n$ or $p\mid a$, the latter is impossible. $\textbf{Remark.}$ As a corollary of this we get that if $q,p$ are primes with $p\mid \Phi_q(a)$ then $p\equiv 1 \pmod q$ or $p=q$. (this is easier to prove) and also gives us immediately that in this case all divisor of $\Phi_q(a)$ is $1\hspace{0.1cm}\text{or} \hspace{0.1cm}0\pmod q$, because all divisor is a product of primes which are $1 \hspace{0.1cm}\text{or} \hspace{0.1cm}0\pmod q$. Solution to the problem The problems follows almost immediately with this, as we need $\Phi_7(x)=(y-1)(y^4+y^3+y^2+y+1)$, so $y-1$ is $1 \hspace{0.1cm}\text{or} \hspace{0.1cm}0\pmod 7$ thus $y^4+y^3+y^2+y+1$ is $5 \hspace{0.1cm}\text{or} \hspace{0.1cm}3\pmod 7$ which is impossible.

Notice the LHS is positive, so we only have to consider $y>0$. It is well known that the prime factors of $\Phi_p$ are either $p$ or $1 \pmod{p}$. Applying this to $p=7$ as the LHS is $\Phi_7$, we see that any prime dividing the RHS follows this format as well. If $y \not \equiv 1 \pmod{7}$, $y-1 \not \equiv 0 \pmod{7}$, so we must instead have $y \equiv 2 \pmod{7}$, which gives $y^4+y^3+y^2+y+1 \equiv 3 \pmod{7}$. If $y \equiv 1 \pmod{7}$, we have $y^4+y^3+y^2+y+1 \equiv 5 \pmod{7}$. Neither of those cases yields $y^4+y^3+y^2+y+1 \equiv 0,1 \pmod{7}$, so there are no solutions. $\square$

I claim that there are no solutions. We first use a small lemma: Lemma: If $p, q$ are primes so that $p \mid \frac{x^q - 1}{x - 1}$, then $q \mid p - 1$ or $q = p$. Proof: Let $o(x)_p$ denote the order of $x$ modulo $p$. Then if $p \mid (x^q - 1)/(x - 1)$ then $p \mid x^q - 1$ so that $o(x)_p \mid q$ and $o(x)_p \mid p - 1$, so that $o(x)_p \mid \gcd(q, p - 1) \in \{1, q\}$. Case 1: $\left(o(x)_p = q \right).$ Then trivially $q \mid p - 1$. Case 2: $\left(o(x)_p = 1 \right).$ Then $p \mid x - 1$, so that $x^{q - 1} + x^{q - 2} + \cdots + x + 1 \equiv q \equiv 0 \pmod p$. Hence $q = p$. $\square$ Now for $q = 7$, we find that $7 \mid p - 1$ or $p = 7$, in particular, every factor of $(x^7 - 1)/(x - 1)$ is either $0, 1$ modulo $7$. Therefore \[y^5 - 1 \equiv 0, 1 \pmod 7 \implies y^5 \equiv 1, 2 \pmod 7 \implies y \equiv 1, 4 \pmod 7. \]However, if $y \equiv 4 \pmod 7$, then $y - 1 \equiv 3 \pmod 7$, which is a contradiction. Similarly if $y \equiv 1 \pmod 7$, then $y^4 + y^3 + y^2 + y + 1 \equiv 5 \pmod 7$, which is another contradiction. Hence no solution exists, as initially claimed. $\blacksquare$

Assume FTSOC that there does exist an integer solution $(x,y)$ to the equation. Note that $(x,y)$ is a solution if and only if $(-x,y)$ is a solution, so we can assume WLOG that $x$ is a positive integer ($x=0$ clearly has no solutions). 1. If a prime $p\neq 7$ divides $\frac{x^7-1}{x-1}$, then $ord_p(x)=7$. Since $7$ is prime, this is basically equivalent to proving that $x$ cannot be $1$ mod $p$ if $p$ divides $\frac{x^7-1}{x-1}$. Assume FTSOC that $x$ is $1$ mod $p$, meaning that $p\mid x-1$. Then by LTE, we have that \[\nu_p(x^7-1)=\nu_p(x-1)+\nu_p(7)=\nu_p(x-1),\]since $p\neq 7$. Therefore, if $p\mid x-1$, that means that $p$ does not divide $\frac{x^7-1}{x-1}$, a contradiction. Therefore we must have that $ord_p(x)=7$. Additionally, extending this claim a bit further, we can also find that since $ord_p(x)=7$, we have that if $p\mid \frac{x^7-1}{x-1}$, then $p$ must be $1$ mod $7$, since $ord_p(x)=7$ must divide $\phi(p)=p-1$. 2. If $7\mid \frac{x^7-1}{x-1}$? Since $7\mid \frac{x^7-1}{x-1}$, we must have that $7\mid x^7-1$. By FLT we have that \[x^7-1 \equiv x-1 \mod 7,\]meaning that in order for $7$ to divide $x^7-1$, we must have that $x$ is $1$ mod $7$. This implies that $7\mid x-1$, and by LTE we have that \[\nu_7(x^7-1)=\nu_7(x-1)+\nu_7(7)=\nu_7(x-1)+1,\]which implies that \[1=\nu_7\left(\frac{x^7-1}{x-1}\right)=\nu_7(y^5-1).\]Using this, we get that we must have $7\mid y^5-1$. Running through all of the mods mod $7$, we get that this means that $y$ must be $1$ mod $7$. Therefore we have \[y^5-1=(y-1)(1+y+y^2+y^3+y^4),\]and since $7\mid y-1$ and $\nu_y(y^5-1)=1$, this means that $7$ cannot divide $1+y+y^2+y^3+y^4$. However, we have from (1) that all prime factors other than $7$ of $\frac{x^7-1}{x-1}=y^5-1$ are $1$ mod $7$, implying that $1+y+y^2+y^3+y^4$ must only have prime factors that are $1$ mod $7$. This means that \[1+y+y^2+y^3+y^4\equiv 1\mod 7 \iff y(1+y+y^2+y^3)\equiv 0\mod 7.\]However, running through all mods mod $7$, we get that $1+y+y^2+y^3$ is never $0$ mod $7$, meaning that we must have $y$ be $0$ mod $7$, a contradiction to our previous claim that $y$ must be $1$ mod $7$. Therefore we have that $7$ cannot divide $\frac{x^7-1}{x-1}$, meaning that this case has no solutions. 3. If $\frac{x^7-1}{x-1}$ is not a multiple of $7$? Since $7$ is not a prime factor of this number, we have that the prime factors of this number must all be $1$ mod $7$ by (1). This means that the number $\frac{x^7-1}{x-1}=y^5-1$ and all its factors must be $1$ mod $7$. Factoring $y^5-1$, we get \[y^5-1=(y-1)(1+y+y^2+y^3+y^4),\]meaning that \[1+y+y^2+y^3+y^4\equiv 1\mod 7 \iff y(1+y+y^2+y^3)\equiv 0\mod 7,\]which we already established in (2) only works if $y$ is $0$ mod $7$. However, this implies that $(y-1)$, a factor of $y^5-1$, is $6$ mod $7$. This is a contradiction to our earlier claim that all factors of $\frac{x^7-1}{x-1}=y^5-1$ are $1$ mod $7$. Therefore there are no solutions in the case if $\frac{x^7-1}{x-1}$ is not a multiple of $7$. Since we have exhausted all possible cases, this implies that there are no integer solutions to this equation, finishing the problem.

We start by noting that other than the illegal case of $x=1$, $\frac{x^7-1}{x-1}$ is always positive. This tells us that $y>1$. If $x=0$ or $x=-1$ then $y^5=2$ is clearly impossible. Thus we have dealt with all of the small cases where no primes divide either side of the equation. We can also show the left side is odd by plugging $0$ and $1$ into $x^6+x^5+\dots+x+1 \pmod{2}$ and seeing that they are both evaluate to $1\pmod{2}$. From here on we only have to worry about odd primes. Consider some odd prime $p$ such that $p\mid x^7-1$. This implies that the order of $x\pmod{p}$ divides $7$, so the order is either $1$ or $7$. If the order is $1$, then $p$ also divides the $x-1$ factor in the denominator. Then by LTE, $\nu_p(x^7-1)=\nu_p(x-1)+\nu_p(7)$. This implies that any $p$ dividing $x-1$ will be canceled in the division, other than $7$, which would have an extra factor in the numerator. If the order is $7$, then $7\mid p-1$ implies that $p\equiv 1\pmod{7}$. This tells us that the left side may only have $7$ or primes $p\equiv1\pmod{7}$ as its prime factors. Now we factor the right side as $(y-1)\left(\frac{y^5-1}{y-1}\right)$. Consider a prime $p$ dividing $\frac{y^5-1}{y-1}$. This works similarly to the left side. If the order of $y$ with respect to $p$ is $1$, then by LTE the factors cancel out unless $p$ is $5$. Otherwise, if the order is $5$, then $p\equiv 1\pmod{5}$. However, we have shown that only $7$ and primes $1\pmod{7}$ can be factors of the left side, so $\frac{y^5-1}{y-1}$ can only contain similar factors. $7$ is not $1\pmod{5}$, so $\frac{y^5-1}{y-1}$ can only contain primes that are $1\pmod{7}$ and thus must be $1\pmod{7}$. We can test values$\pmod{7}$ to see that $\frac{y^5-1}{y-1}\equiv 1\pmod{7}$ only occurs when $y\equiv 0\pmod{7}$ or $y\equiv{6}\pmod{7}$. However, in both of these cases, $y-1$ is not $0$ or $1\pmod{7}$, and thus contains prime factors that the left side does not, contradiction. Thus there are no integer solutions to the equation, as desired.

We claim there are no solutions. Suppose $(x,y)$ is an integer solution to the equation. Then, consider a prime $p$ dividing $\frac{x^7 - 1}{x-1} = y^5 - 1$. By the divisors of cyclotomic polynomials lemma, $p = 7$ or $p \equiv 1 \pmod{7}$. This means every factor of $y^5 - 1$ is either $0$ or $1$ mod $7$. Consider the case where $y^5 - 1 \equiv 0 \pmod{7}$. Then, we have $y \equiv 1 \pmod{7}$. Since $y^4 + y^3 + y^2 + y^1 + 1 \equiv 5 \pmod{7}$ is a factor dividing $y^5 - 1$, we have a contradiction. If $y^5 - 1 \equiv 1 \pmod{7}$, then we have $y \equiv 4 \pmod{7}$. But then, $y-1 \equiv 3 \pmod{7}$ is a factor dividing $y^5 - 1$, so we have a contradiction. Thus, there are no solutions.

For any prime factor $p \mid \frac{x^7-1}{x-1}$ we have that $p \mid x^7-1$, which implies $x^7 \equiv 1 \mod{p}$, so $\delta \mid \gcd(7,p-1) \in \{1,7\}$. If $\delta=1$ then $x \equiv 1 \mod{p} \implies x^6+\cdots+1 \equiv 7 \equiv 0 \mod{p} \implies p=7$. If $\delta=7$ then $7 \mid p-1 \implies p \equiv 1 \mod{7}$. So all prime factors of $\frac{x^7-1}{x-1}$ are either $7$ or $1 \mod{7}$. That means $y^5-1=(y-1)(y^4+y^3+y^2+y+1)$ must only have prime factors that are $7$ or $1 \mod{7}$. If $y-1 \equiv 0 \mod{7}$ then $y^4+\cdots+1 \equiv 5 \mod{7}$ which means it must contain a prime factor that doesn't belong to the LHS. Thus $y-1$ cannot contain $7$ as a prime factor. So suppose now all its prime factors are $1 \mod{7}$, then $y \equiv 2 \mod{7}$ so $y^4+\cdots+1 \equiv 31 \equiv 3 \mod{7}$ which means it contains a prime factor not belonging to the LHS. Thus this equation has no integer solutions.

shane wrote: if we change 7 by 3,can we have the same solution? No, because then in fact a solution does exist: \[ \frac{5^3-1}{5-1} = 31 = 2^5-1. \]

Consider a prime $p$ that divides $y^5 -1$, then $p$ divides $x^7 -1$ implying order of $p$ is $1$ or $7$ implying all prime factors of $LHS$ are $7$ or $1(mod \ 7)$ so $y^5 -1$ must be $0$ or $1$ mod $7$. Now if $7$ divides $y-1$ then $RHS$ divided by $y-1$ is $5$ modulo $7$ which cannot be. But if $y^5 -1$ is $1$ modulo $7$ then $y \equiv 4 (mod \ 7)$ implying $y-1 \equiv 3(mod \ 7)$ which is again a contradiction hence no solutions.

Consider some prime factor $p\mid\frac{x^7-1}{x-1}$. Then $x^7\equiv x^{p-1}\equiv1\pmod{p}$ and hence $x^{\gcd(p-1,7)}\equiv1\pmod{p}$. If $\gcd(p-1,7)=7$, then $p\equiv1\pmod7$. Otherwise, we have $x\equiv1\pmod{p}$ and thus $p\equiv0\pmod7$. Hence the factors of $y^5-1$ are all either $0$ modulo $7$ or $1$ modulo $7$. If $y\equiv1\pmod{7}$, then $y^4+y^3+y^2+y+1\equiv5\pmod7$ is a factor of $y^5-1$, a contradiction. If $y\equiv2\pmod{7}$, then $y^5-1\equiv3\pmod7$. Otherwise, note that $y-1$ is a factor of $y^5-1$ but is not $0$ or $1$ modulo $7$, another contradiction. We are done. $\blacksquare$

We claim that there are no solutions: For the sake of contradiction, assume there exists solutions. Then note that: $\frac{x^5-1}{x-1}=y^5-1$ cant have the value of $\pm 1$ for integers $x, y$. Consider a prime $p$ which divides $\tfrac{x^7-1}{x-1}$. Then: $\text{ord}_p(x)=1, 7$. If $\text{ord}_p(x)=1$, then $x \equiv 1 \pmod p$ and thus: $\tfrac{x^7-1}{x-1} \equiv x^6+x^5+\cdots+x+1 \equiv 7 \pmod p$. Thus: $p=7$. If $\text{ord}_p(x)=7$, then $7|p-1$ or $p \equiv 1 \pmod 7$. Thus: every prime factor is either $p=7$ or $p \equiv 1 \pmod 7$. Consider $y^5-1=(y-1)(y^4+y^3+y^2+y+1)$. Each factor is either $0, 1 \pmod 7$. If $y-1 \equiv 0 \pmod 7$, then: $y^4+y^3+y^2+y+1 \equiv 5 \pmod 7$ which leads to contradiction. If $y-1 \equiv 1 \pmod 7$, then: $y^4+y^3+y^2+y+1 \equiv 3 \pmod 7$ which leads to contradiction. Hence, there are no solutions.

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