We use the fact that if $n\in\mathbf{Z}^+,x\in\mathbf{Z}$, then for all primes $p|\Phi_n(x)$, we must have either $p|n$ or $p\equiv 1\pmod{n}$. Applying this to $\Phi_p(x)$ for some prime $p$, we get that every prime $q|\Phi_p(x)$ satisfies either $q=p$ or $q\equiv 1\pmod{p}$. We can rewrite the given as $\Phi_7(x)=(y-1)\Phi_5(y)$ .Suppose $p|\Phi_7(x)$. Then either $p=7$ or $p$ is of the shape $7k+1$. Thus every divisor of $\Phi_7(x)$ is either $0$ or $1$ modulo 7, so $(y-1)$ is 0 or 1 modulo 7, so $y$ is 1 or 2 modulo 7. In the former case, $\Phi_5(y)\equiv 5\pmod{7}$, contradiction, and in the latter case $\Phi_5(y)\equiv 3\pmod{7}$, contradiction, so there are no integer solutions. EDIT: facepalm

tc1729 wrote: We can rewrite the given as $\Phi_7(x)=(y-1)\Phi_5(y)$ It should be $y+1$, which wrecks this proof. This problem is not simply 2006 ISL N5. Re-write the equation as $x(x^5 + x^4 + x^3 + x^2 + x + 1) = y^5$. Remark that $\gcd(x,x^5 + x^4 + x^3 + x^2 + x + 1) = 1$ so it follows $x^5 + x^4 + x^3 + x^2 + x + 1$ is a power of $5$. However for $x > 1$ we have $x^5 < x^5 + x^4 + x^3 + x^2 + x + 1 < (x+1)^5$, so it follows no solutions exist.

Yeah . Or simply factorize , $x(x(x^4+x^2+1)+x^4+x^2+1)) = x(x+1)(x^4+x^2+1)=y^5$. This means that both of $x+1,x$ are 5th power , which is contradiction. tc1729 : I also thought that they copied ISL problem , when i saw it for the 1st time!

$(x^7-1)/(x-1)=x^6+x^5+x^2+x+1$ than $y^5+1=x^6+x^5+x^2+x+1$ $\Rightarrow$ $y^5=x(x+1)(x^2+x+1)$. Than those equation has no natural solution.

SamISI1 wrote: $(x^7-1)/(x-1)=x^6+x^5+x^2+x+1$ than $y^5+1=x^6+x^5+x^2+x+1$ $\Rightarrow$ $y^5=x(x+1)(x^2+x+1)$. Than those equation has no natural solution. This is wrong.. $\frac{x^7-1}{x-1}\neq{x^6+x^5+x^2+x+1}$ It is, $(x^6+x^5+x^4+x^3+x^2+x+1)$ $\because$ $(x^7-1)=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.. Hope you get your mistake

Hello this is mine solution : \frac{x^{7}-1}{x-1} =( x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x^{1} + 1) which means y^{5} = x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x^{1} which means y^{5} = x^{2}( x^{4} + x^{3} + x^{2} + x^{1} + 1) which means (x^{4} + x^{3} + x^{2} + x^{1} + 1) is of form x^{3}*q^{5} for some q\in{\mathbb{N}} which get y = xq but from this to happen x^{3}\mid{(x^{4} + x^{3} + x^{2} + x^{1} + 1)} which means x^{3}\mid{(x^{2}+x^{1}+1)} which means x^{3}\mid{\frac{x^{3}-1}{x-1}} which means \frac{x^{3}-1}{x-1}\equiv{0}(mod x^{3}) but note that gcd(x^3,x-1)=1 { (PROOF:gcd(x^3,x-1)=gcd(x^2,x-1)=gcd(1,x-1)=1)} which means \frac{x^{3}-1}{x-1}\equiv{(x^{3}-1)({x-1}^{-1})}\equiv{(-1)({x-1}^{-1})}\equiv{0} taken mod(x^{3}) which false hence no solution @below thanks ... actually i am new to AOPS due to which i cannot use $$

PSS_73939133 wrote: Hello this is mine solution : $\frac{x^{7}-1}{x-1} =( x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x^{1} + 1) \implies y^{5} = x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x^{1} \implies y^{5} = x^{2}( x^{4} + x^{3} + x^{2} + x^{1} + 1) \implies (x^{4} + x^{3} + x^{2} + x^{1} + 1)$ is of form $x^{3}*q^{5}$ for some $q\in{\mathbb{N}}$ which get $y = xq$ but from this to happen $x^{3}\mid{(x^{4} + x^{3} + x^{2} + x^{1} + 1)} \implies x^{3}\mid{(x^{2}+x^{1}+1)}$ which means $x^{3}\mid{\frac{x^{3}-1}{x-1}} \implies \frac{x^{3}-1}{x-1}\equiv{0}(mod x^{3}).$ but note that $gcd(x^3,x-1)=1$ (PROOF:$gcd(x^3,x-1)=gcd(x^2,x-1)=gcd(1,x-1)=1)$ which means $\frac{x^{3}-1}{x-1}\equiv{(x^{3}-1)({x-1}^{-1})}\equiv{(-1)({x-1}^{-1})}\equiv{0}$ taken $mod(x^{3}$ which false hence no solution idek if i fixed that for you but that gave me a headache