Solve in positive integers: \[x^{y^2+1}+y^{x^2+1}=2^z\]
Problem
Source: 2024 Israel TST Test 2 P1
Tags: number theory, Diophantine equation, Exponential equation
07.11.2023 13:51
If $x=1$ equation will be $y^2+1=2^z$ and $(mod 4)$ give us $(x,y,z)=(1,1,1)$ Now $x,y\ge 2$ If $x,y$ are odd numbers $(mod 4)$ give us contradaction. And for the case $x,y$ is even. I wrote as $x=2^k.x_1$ and $y=2^k.y_1$ where WLOG $x_1$ is odd. If $y_1$ is odd too from $v_2(x^{y^2+1})=v_2(y^{x^2+1})$ we see that $x=y=2^t$ and $(x,y,z)=(2^t,2^t,t(2^{2t}+1)+1)$ for all $t \in Z^+$ If $y_1$ is even we write $y_1=2^{t.2^{2k}}.m_1$ and the problem is straightforward from $v_2(x^{y^2+1})=v_2(y^{x^2+1})$ and we get contradaction.
22.11.2023 10:48
LuciferMichelson wrote: If $y_1$ is even we write $y_1=2^{t.2^{2k}}.m_1$ and the problem is straightforward from $v_2(x^{y^2+1})=v_2(y^{x^2+1})$ and we get contradaction. Sorry for my stupidity, but how can you able to archive that form of $y_1$? I tried to prove this case but it seems quite difficult.
22.11.2023 11:38
From $v_2(x^{y^2+1})=v_2(y^{x^2+1})$ we get $2^{2k}k({y_1}^2-{x_1}^2)=v_2(y_1).(2^{2k}.{x_1}^2+1)$ so from $(2^{2k},2^{2k}.{x_1}^2+1)=1$ we get $2^{2k}|v_2(y_1)$
22.11.2023 12:00
LuciferMichelson wrote: From $v_2(x^{y^2+1})=v_2(y^{x^2+1})$ we get $2^{2k}k({y_1}^2-{x_1}^2)=v_2(y_1).(2^{2k}.{x_1}^2+1)$ so from $(2^{2k},2^{2k}.{x_1}^2+1)=1$ we get $2^{2k}|v_2(y_1)$ Got it now. Much appreciated!
03.02.2024 17:01
My Problem! As the RHS is even, $x,y$ must have the same parity. If they are odd we get that the LHS is $1 \pmod 4$, which means that $z=1$, i.e. $$x^{y^2+1}+y^{x^2+1}=2 \leq x+y \implies x=y=z=1$$Assume now that $x,y$ are even. if $v_2(x^{y^2+1}) < v_2(y^{x^2+1})$ we get that $v_2(x^{y^2+1})=v_2(x^{y^2+1}+y^{x^2+1})=v_2(2^z)=z \implies 2^z \mid x^{y^2+1}$, which is a contradiction, as $x^{y^2+1}<x^{y^2+1}+y^{x^2+1}=2^z$. Similarly, we get a contradiction in the case $v_2(x^{y^2+1}) > v_2(y^{x^2+1})$. Therefore: $$(y^2+1)v_2(x)=v_2(x^{y^2+1}) = v_2(y^{x^2+1})=(x^2+1)v_2(y)$$Notice that if $gcd(x^2+1,y^2+1)=1$ we get that $x^2+1 \mid v_2(x) < x < x^2+1$, which is a contradiction, and therefore $gcd(x^2+1,y^2+1)=d>1$. We can now write $a=x^{\frac{y^2+1}{d}},b=y^{\frac{x^2+1}{d}}$ to get $a^d+b^d=2^z$. Denote $gcd(a,b)$ by $c$ and write $a=a'c,b=b'c$. Notice that $c$ must be a power of $2$. Then $(a')^{d}+(b')^{d}=2^{z'}$ and $gcd(a',b')=1$, but by LTE we know that $v_2((a')^{d}+(b')^{d})\leq v_2(a+b)$, which means that $a'=b'=1$ (as we know that $d>1$). Therefore, $a=b=2^{\frac{z-1}{d}}$ which means that $x^{\frac{y^2+1}{d}}=y^{\frac{x^2+1}{d}}=2^{\frac{z-1}{d}}$. Therefore $x^{\frac{1}{x^2+1}}=y^{\frac{1}{y^2+1}}$, and it's easy to see it means that $x=y$. By looking again at the original equation we get that $x=y=2^t$ for some $t$, which gives the solution $(2^t,2^t,t2^{2t}+t+1)$.
20.02.2024 14:55
Unusually cool for a Diophantine equation. Great problem.
20.06.2024 15:25
I initially freaked out on this but I think I got it now. We claim that the answer is all triples of non-negative integers $(2^r,2^r,r(2^r+1)+1)$. It is easy to see that these triples indeed work. Now, we show that they are the only ones. Since the right hand side is even for all positive integers $z$, it immediately follows that $x$ and $y$ are of the same parity. Thus, we have two cases. Case 1 : $x \equiv y \equiv 1 \pmod{2}$. But then, $x^2+1 \equiv y^2+1 \equiv 0 \pmod{2}$ so we have that, \[x^{y^2+1} + y^{x^2+1} \equiv 1 + 1 \equiv 2 \pmod{4}\]which implies that $z=1$ and in turn, $x=y=1$. Case 2 : Now, we let $x=a^2l$ and $y=2^bm$ for positive integers $a,b,l,m$ where $l$ and $m$ are both odd. Throughout this proof we use the result that if for two even integers we have $n_1+n_2=2^{i}$ for some positive integer $i$, then $\nu_2(n_1)=\nu_2(n_2)$. This is clear since otherwise, factoring out the common power of two, we are left with an odd factor greater than 1. Thus, \begin{align*} (2^al)^{y^2+1} + (2^bm)^{x^2+1} &= 2^z\\ 2^{a(y^2+1)} l^{y^2+1} + 2^{b(x^2+1)}m^{x^2+1} &= 2^z \end{align*}Now, by the previously mentioned result, we must have $a(y^2+1)=b(x^2+1)=s$. Thus, \[l^{y^2+1}+m^{x^2+1}=2^{z-s}\]Now, if $z>s+1$, the right hand side is clearly divisible by 4. But, since $y^2+1$ and $x^2+1$ are both odd, $l^{y^2+1}+m^{x^2+1} \equiv l +m \pmod{4}$. Thus, $l+m$ is divisivle by 4. Since $l$ and $m$ are both odd, it follows that one of them is $1\pmod{4}$ and the other is $3\pmod{4}$. WLOG, we assume that $l \equiv 1 \pmod{4}$ and $m\equiv 3 \pmod{4}$. Note that, \[2^{z-s} = (l^{y^2+1}-1)+(m^2{x^2+1}+1)\]But by LTE, \[\nu_2(l^{y^2+1}-1)=\nu_2(l-1)+\nu_2(l+1)+\nu_2(y^2+1)-1 = \nu_2(l-1)\]and $\nu_2(m^{x^2+1}+1) = \nu_2(m+1)$. Thus, we must have $\nu_2(l-1)=\nu_2(m+1)$. Similarly, \[2^{z-s} = (l^{y^2+1}+1)+(m^2{x^2+1}-1)\]But applying LTE again, $\nu_2(l^{y^2+1}+1)=\nu_2(l+1)$ and \[\nu_2(m^{x^2+1}-1)=\nu_2(m-1)+\nu_2(m+1)+\nu_2(x^2+1)-1=\nu_2(m+1)\]Thus, $\nu_2(l+1)= \nu_2(m+1)$. But this implies that, \[\nu_2(l+1)= \nu_2(m+1)=\nu_2(l-1)\]which is a clear contradiction for all positive integers $l$. Thus, we have only two possibilities. When $z=s$, \[2^s(l^{y^2+1}+m^{x^2+1})=2^s\]so, $l^{y^2+1}+m^{x^2+1}=1$ which has no solutions in the positive integers. When $z=s+1$, \[2^s(l^{y^2+1}+m^{x^2+1})=2^{s+1}\]so $l^{y^2+1}+m^{x^2+1}=2$ which implies that $l=m=1$. Thus, $x=2^a$ and $y=2^b$. Since, we have from before that, \begin{align*} a(y^2+1) &= b(x^2+1)\\ a(2^b+1) &= b(2^a+1)\\ \frac{2^a+1}{a} &= \frac{2^b+1}{b} \end{align*}which implies that $a=b$. Thus, we must have $x=y=2^a$. Plugging in these values for $x$ and $y$ implies that $z=a(2^a+1)+1$ as expected. Thus, we have exhausted all solutions and indeed all solutions are of the claimed forms.
02.07.2024 14:45
cursed_tangent1434 wrote: I initially freaked out on this but I think I got it now. We claim that the answer is all triples of non-negative integers $(2^r,2^r,r(2^r+1)+1)$. It is easy to see that these triples indeed work. Now, we show that they are the only ones. Since the right hand side is even for all positive integers $z$, it immediately follows that $x$ and $y$ are of the same parity. Thus, we have two cases. Case 1 : $x \equiv y \equiv 1 \pmod{2}$. But then, $x^2+1 \equiv y^2+1 \equiv 0 \pmod{2}$ so we have that, \[x^{y^2+1} + y^{x^2+1} \equiv 1 + 1 \equiv 2 \pmod{4}\]which implies that $z=1$ and in turn, $x=y=1$. Case 2 : Now, we let $x=a^2l$ and $y=2^bm$ for positive integers $a,b,l,m$ where $l$ and $m$ are both odd. Throughout this proof we use the result that if for two even integers we have $n_1+n_2=2^{i}$ for some positive integer $i$, then $\nu_2(n_1)=\nu_2(n_2)$. This is clear since otherwise, factoring out the common power of two, we are left with an odd factor greater than 1. Thus, \begin{align*} (2^al)^{y^2+1} + (2^bm)^{x^2+1} &= 2^z\\ 2^{a(y^2+1)} l^{y^2+1} + 2^{b(x^2+1)}m^{x^2+1} &= 2^z \end{align*}Now, by the previously mentioned result, we must have $a(y^2+1)=b(x^2+1)=s$. Thus, \[l^{y^2+1}+m^{x^2+1}=2^{z-s}\]Now, if $z>s+1$, the right hand side is clearly divisible by 4. But, since $y^2+1$ and $x^2+1$ are both odd, $l^{y^2+1}+m^{x^2+1} \equiv l +m \pmod{4}$. Thus, $l+m$ is divisivle by 4. Since $l$ and $m$ are both odd, it follows that one of them is $1\pmod{4}$ and the other is $3\pmod{4}$. WLOG, we assume that $l \equiv 1 \pmod{4}$ and $m\equiv 3 \pmod{4}$. Note that, \[2^{z-s} = (l^{y^2+1}-1)+(m^2{x^2+1}+1)\]But by LTE, \[\nu_2(l^{y^2+1}-1)=\nu_2(l-1)+\nu_2(l+1)+\nu_2(y^2+1)-1 = \nu_2(l-1)\]and $\nu_2(m^{x^2+1}+1) = \nu_2(m+1)$. Thus, we must have $\nu_2(l-1)=\nu_2(m+1)$. Similarly, \[2^{z-s} = (l^{y^2+1}+1)+(m^2{x^2+1}-1)\]But applying LTE again, $\nu_2(l^{y^2+1}+1)=\nu_2(l+1)$ and \[\nu_2(m^{x^2+1}-1)=\nu_2(m-1)+\nu_2(m+1)+\nu_2(x^2+1)-1=\nu_2(m+1)\]Thus, $\nu_2(l+1)= \nu_2(m+1)$. But this implies that, \[\nu_2(l+1)= \nu_2(m+1)=\nu_2(l-1)\]which is a clear contradiction for all positive integers $l$. Thus, we have only two possibilities. When $z=s$, \[2^s(l^{y^2+1}+m^{x^2+1})=2^s\]so, $l^{y^2+1}+m^{x^2+1}=1$ which has no solutions in the positive integers. When $z=s+1$, \[2^s(l^{y^2+1}+m^{x^2+1})=2^{s+1}\]so $l^{y^2+1}+m^{x^2+1}=2$ which implies that $l=m=1$. Thus, $x=2^a$ and $y=2^b$. Since, we have from before that, \begin{align*} a(y^2+1) &= b(x^2+1)\\ a(2^b+1) &= b(2^a+1)\\ \frac{2^a+1}{a} &= \frac{2^b+1}{b} \end{align*}which implies that $a=b$. Thus, we must have $x=y=2^a$. Plugging in these values for $x$ and $y$ implies that $z=a(2^a+1)+1$ as expected. Thus, we have exhausted all solutions and indeed all solutions are of the claimed forms. why is v_2 (m^{x^2+1}-1) that? and not v_2 (m-1)