Observe that by letting $n = 2^k$ in the equation, we know that $2^{2k}y^2 + 1| x^{2^{k-1}}-1 \Rightarrow 4^ky^2 + 1| x^{4^k}-1$. However, this is just https://artofproblemsolving.com/community/c6h545765p3156844 (ISL 2012 N6) with $2$ replaced by $4$. This is not hard to do, below is an adaptation of post #19 in the linked thread. It suffices only to prove that if $a, x \in \mathbb{N}$ are such that $4^ka + 1| x^{4^k} -1$ for all $k \in \mathbb{N}$, then $x = 1$. Take a power of two $2^{2z} >>> 4a+1.$ Call a prime $p$ chaotic if $p \equiv 1$ (mod $2^{2z}$) does not hold. We claim that there are infinitely many chaotic primes dividing $4^ka+1$ for some $k \in \mathbb{N}$, which would solve the problem since they would all have to divide $x^{2^{2z-1}} - 1$, which clearly gives $x = 1.$ To show this, we assume the contrary. Suppose that only $p_1, p_2, \cdots, p_c$ are the only chaotic primes which divide $4^ka+1$ for some $k \in \mathbb{N}.$ Let $P = p_1p_2 \cdots p_c.$ Now, observe that in modulo $2^{2z}$ we have that for all $t$ sufficiently large $$2^{2 + 2zP^t\phi(P)}a + 1 \equiv 1,$$and that $2^2a+1 \equiv 4a + 1$. If we take $a$ to be larger than $v_{p_i}(4a+1)$ for $i = 1, 2, \cdots, c$, then we see that $2^{2+2zP^t\phi(P)}a + 1$ and $4a+1$ have exactly the same chaotic prime divisors (multiplicity as well). Therefore, they must be congruent mod $2^{2z}$, since any non-chaotic prime divisor doesn't change (mod $2^{2z}$) value. As a result, we've that $4a + 1 \equiv 1$, clearly absurd since $2^{2z} >>> 4k+1.$ $\square$

Take a big prime $p\equiv 5\pmod{12},$ and let $b$ be such $b\equiv iy^{-1} \pmod{p}.$ Let $n=kp(p-1)-p(b-3)+b$ be a prime, then $1+(ny)^2 \equiv 0 \pmod{p}$ which means that $p \mid x^{\varphi(n)}-1\implies \text{ord}_p(x) \mid \gcd(\varphi(n), p-1),$ but since $n-1 \equiv 2 \pmod{p-1},$ we have that $p \mid x^2-1,$ clearly a contradiction unless $x=1.$

rmtf1111 wrote: Take a big prime $p,$ and let $b$ be such $b\equiv iy^{-1} \pmod{p}.$ Let $n=kp(p-1)-p(b-3)+b$ be a prime, then $1+(ny)^2 \equiv 0 \pmod{p}$ which means that $p \mid x^{\varphi(n)}-1\implies \text{ord}_p(x) \mid \gcd(\varphi(n), p-1),$ but since $n-1 \equiv 2 \pmod{p-1},$ we have that $p \mid x^2-1,$ clearly a contradiction unless $x=1.$ Sorry if this is a dumb question but doesn't $b\equiv iy^{-1} \pmod{p}.$ make the remainder complex ? Or since all integers are complex so this is fine?

Pluto1708 wrote: rmtf1111 wrote: Take a big prime $p,$ and let $b$ be such $b\equiv iy^{-1} \pmod{p}.$ Let $n=kp(p-1)-p(b-3)+b$ be a prime, then $1+(ny)^2 \equiv 0 \pmod{p}$ which means that $p \mid x^{\varphi(n)}-1\implies \text{ord}_p(x) \mid \gcd(\varphi(n), p-1),$ but since $n-1 \equiv 2 \pmod{p-1},$ we have that $p \mid x^2-1,$ clearly a contradiction unless $x=1.$ Sorry if this is a dumb question but doesn't $b\equiv iy^{-1} \pmod{p}.$ make the remainder complex ? Or since all integers are complex so this is fine? In the reals, $i$ is defined as the number $x$ such that $x^2 = -1$. Similarly, in modulos you could define $i$ is the number such that $x^2 \equiv -1 \pmod k$ for some integer $k$. People with a good abstract algebra understanding will instantly know what he's talking about. OR, he could've used it as a constant or variable

Pluto1708 wrote: rmtf1111 wrote: Take a big prime $p,$ and let $b$ be such $b\equiv iy^{-1} \pmod{p}.$ Let $n=kp(p-1)-p(b-3)+b$ be a prime, then $1+(ny)^2 \equiv 0 \pmod{p}$ which means that $p \mid x^{\varphi(n)}-1\implies \text{ord}_p(x) \mid \gcd(\varphi(n), p-1),$ but since $n-1 \equiv 2 \pmod{p-1},$ we have that $p \mid x^2-1,$ clearly a contradiction unless $x=1.$ Sorry if this is a dumb question but doesn't $b\equiv iy^{-1} \pmod{p}.$ make the remainder complex ? Or since all integers are complex so this is fine? A way to think about it is that since $p \equiv 1 \pmod{4}$, there exists $x$ such that $p \mid x^2+1$ (quick proof using primitive roots), but now this is just $x^2=-1$ in $\mathbb{F}_p$. The $i$ is just a way to denote the residue class of this $x$ in $\mathbb{F}_p$.

Thanks a lot @above

We start off with a claim : Claim : There are infinitely many primes of the form $5\pmod {12}$ dividing atleast one term of the sequence $\{a_n\}_{n\geq 0}$ given by : $$a_n = \left( 3^n \cdot y \right)^2+1$$ Proof : Assume to the contrary that there are only finitely many such primes. Since any prime divisor of the sequence is $1\pmod 4$, hence there are only finitely many prime divisors of the sequence of the form $2\pmod 3$. Suppose the product of those primes is $P$. WLOG assume that $3\nmid y$ , so that $y ^2+1 \equiv 2 \pmod 3$ has a $2\pmod 3$ prime divisor. Consider $N=(y^2+1)\cdot P$. We have $$a_{\varphi(N)} = \left( 3^{\varphi (N)} \cdot y \right)^2 \equiv y^2+1 \pmod N$$ Hence $a_{\varphi (N)}= (y^2+1)\cdot (xP+1)$ for some $x$. Note that since $a_{\bullet}\equiv 1 \pmod3$ and $y ^2+1 \equiv 2 \pmod 3$ , hence $(xP+1)$ must have a $2\pmod 3$ prime factor which doesnt divide $P$. Contradiction.$\blacksquare$ Now pick a large prime $q\equiv 5\pmod {12}$ such that $q\mid \left( 3^n \cdot y \right)^2+1$ for some $n$. Suppose $d$ is the order of $x\pmod q$. Obviously $q-1 \nmid 3^{n-1}$ , so $d\mid 2$. Hence $q\mid x^2-1$. Choose $q$ to be sufficiently large to get $x^2=1 \rightarrow \boxed {x=1}$. $\blacksquare$

Consider a prime $p > x^2$ such that $p\equiv 1 \pmod 4$ and $\gcd(p,x)=\gcd(p,y)=\gcd(p,y^2+1)=1$. Define a positive integer $m$ less than $p$ such that $p\mid (my)^2+1$. (There must exist such $m$ because $p\equiv 1\pmod 4$) From $\gcd(p,y^2+1)=1$, we have $m\neq 1$. Then, we have that for every positive integer $n\equiv m\pmod p$, $x^{\phi(n)}-1$ is divisible by $p$. By dirichlet, there exist a prime $q$ such that $q\equiv m\pmod p$. Because $m\neq 1$, we have $\gcd(p,q-1)=1$. By dirichlet again, there exist a prime $r$ such that $r\equiv m\pmod p$ and $r\equiv q-2\pmod {q-1}$. Because $q$ and $r$ are prime, we have $\phi(q)=q-1$ and $\phi(r)=r-1$ $p$ is divisible by $x^{q-1}-1$ and $x^{r-1}-1$. So, $p\mid x^{\gcd(q-1,r-1)}-1$. From $r\equiv q-2\pmod {q-1}$, we conclude that $\gcd(q-1,r-1)$ must divide 2. So, $p\mid x^2-1$. However, $p>x^2$ implies that $x^2-1=0$. Therefore, $x=1$.

Nice Problem !! First of all take an odd prime $p | x -1 $ , we have- $$ (ny)^2 + 1 | x^{\phi(n)} - 1 $$, so we must have $$v_p ( x^{\phi(n)} - 1) \ge v_p ( (ny)^2 +1) $$By LTE , we must have $ v_p ( x-1) + v_p(\phi(n)) \ge v_p ( (ny)^2 + 1) $ , for all natural $n$ , So if we can construct a $n$ such that $ v_p ( x-1) + v_p(\phi(n)) < v_p ( (ny)^2 + 1) $ , then we will get desired contradiction.. Claim : There exist a $n$ such that $ (ny)^2 + 1 \equiv 0 ( mod p^{\alpha}) $ , where $\alpha >>>> v_p(x-1)$. Proof - We want $ n^2 y^2 \equiv -1 ( mod p^{\alpha}) $ , as there exist a primitive root modulo $ p^ {\alpha}$ , Let $g$ be a primitive root mod $ p^ {\alpha} $ , Let $y \equiv g^k ( mod p^ {\alpha})$ , and $ n \equiv g^l (mod p^ {\alpha}) $ , where $k$ is fixed and $l$ is a degree of freedom , Substituting we should get $$ g ^ {2l + 2k} \equiv -1 ( mod p^ {\alpha} )$$, So we can get a unique $l$ $ ( mod \phi(p^ {\alpha}))$. $\blacksquare$. Now suppose from above claim we got $ n \equiv \beta ( mod p^ {\alpha} )$ , so let $$ n = kp^ {\alpha} + \beta $$, now $k$ is a degree of freedom . Note that we have to ensure that $ v_p(\phi(n))$ is not that much large so it violates the condition $ v_p ( x-1) + v_p(\phi(n)) < v_p ( (ny)^2 + 1) $ , So it would be best if we can make $ v_p(\phi(n))$ to be zero , by choosing such a $k$. It can be ensured when every prime factor of $n$ is less than $p$ , So here for constructing a $k$ , we use CRT.. Consider the system of congruences - $$\begin{cases} n = kp^ {\alpha} + \beta \equiv x_1 ( mod p_1) \\ kp^ {\alpha} + \beta \equiv x_2 ( mod p_2)\\.\\.\\.\\.\\.\\.\\.\\ kp^ {\alpha} + \beta \equiv x_v ( mod p_v ) \end{cases}. $$ Where $ p_1, p_2 , p_3 ,.... p_v $ are all primes $ > p $ and less than $ n$ and none of $ x_i$ is $0 mod p_i$ . This has a solution by CRT in $k$ $ mod P$ , where $P = \prod p_i $. And from this we can ensure that every prime factor of $n$ is $ < p $. So we should have $x^{\phi(n)} - 1 = 0 $ $\rightarrow$ $ x = 1$ Hence we are done . $\blacksquare$

XbenX wrote: Let $x,y$ be positive integers. If for each positive integer $n$ we have that $$(ny)^2+1\mid x^{\varphi(n)}-1.$$Prove that $x=1$. (Silouanos Brazitikos, Greece) Claim:- There are infinitely many primes of the form $5 \pmod {12 }$ dividing atleast one term of the sequence $\{a_n\}_{n\geq 0}$ given by : $$a_n = \left( 3^n \cdot y \right)^2+1$$Proof:- FTSOC,the set of prime factors of the form $5 \pmod {12}$ are finite given by $$\mathcal{P}=\{p_1,p_2,p_3,........,p_n\}$$By CRT choose $\mathcal{M}$ such that \begin{align*} \mathcal{M}^2+1 \equiv 0 \pmod {p_1^{2e_1}} \\.\\.\\.\\.\\.\\.\\.\\.\\.\\.\\ \mathcal{M}^2+1 \equiv 0 \pmod {p_n^{2e_n}} \end{align*}with $\mathcal{M}^2+1>y$ Then $$a_{k \cdot \phi(\mathcal{M}^2+1)} \equiv k \cdot y^2+1 \pmod {p_i^{2e_1}}$$for $p_i>y^2+1$(which exists otherwise we look at the exponents for a contradiction) Obviously by choosing \begin{align*} y^2 \equiv x \pmod p \\ k \equiv \frac{-1}{x} \pmod p \end{align*}we get the result. Now pick a large prime $q\equiv 5\pmod {12}$ such that $q\mid \left( 3^n \cdot y \right)^2+1$ for some $n$. Suppose $d$ is the order of $x\pmod q$. Obviously $q-1 \nmid 3^{n-1}$ , so $d\mid 2$. Hence $q\mid x^2-1$. Choose $q$ to be sufficiently large to get $x^2=1 \rightarrow \boxed {x=1}$. $\blacksquare$

XbenX wrote: Let $x,y$ be positive integers. If for each positive integer $n$ we have that $$(ny)^2+1\mid x^{\varphi(n)}-1.$$Prove that $x=1$. (Silouanos Brazitikos, Greece) Take a big prime $p=1(mod4)$ then the equation $m^2=-1(modp)$ has sollution. Select now $n$ such that $n=m/y(modp)$ and $n=2(modp-1)$ by Chinese remainder theorem this is similar to $n=a(mod p(p-1))$. Now if we take $n$ such that is also a prime number (we can do this by Deleclert) we gave that: $p|(ny)^2+1|x^{\varphi(n)}-1$ but $|x^{\varphi(n)}-1=x^{n-1}-1=x-1(modp)$ so we have that $x=1$