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Posted: **Mon May 24, 2021 11:46 am**

Find all positive integers $x,y,z$ such that $3^x +4^y=5^z$
Disclaimer:
Don't just say done by Fermat's Last theorem

Posted: **Mon May 24, 2021 10:22 pm**

IDK how it can be done with Fermat's last Theorem, allow me to use Catalan's conjecture instead.

Let me state this theorem (it's proven, but the name stayed like this anyway!),
Catalan's conjecture : If $x,y,m,n$ are positive integers where $m\geq2, n\geq2$, and they satisfies \[x^m-y^n=1\] then $(x,y,m,n)=(3,2,2,3)$.

Now solution to our problem :
Let's start with taking $\pmod{3}$,
$3^x+4^y=5^z$
$\Rightarrow 1\equiv(-1)^z\pmod{3}$
$\Rightarrow z=2c$ for some positive integer $c$.

Let's take $\pmod{4}$,
$(-1)^x\equiv1\pmod{4}$
$\Rightarrow x=2a$ for some positive integer $a$.

Let's manipulate the equation,
$2^{2y}=5^{2c}-3^{2a}$

Case 1 : $c=1$.
In this case, we get $4^y=25-9^a$ and the only solution is, $a=1, y=2$. Because when $a\geq2, 25-9^a<0$.
So, this case gives us the solution $(x,y,z)=(2,2,2)$.

Case 2 : $c\geq2$.
$2^{2y}=(5^c-3^a)(5^c+3^a)$
Let's assume,
$5^c-3^a=2^u\cdots\cdots\cdots(1)$
$5^c+3^a=2^v\cdots\cdots\cdots(2)$
Where $u,v$ are nonnegative integers, $v>u$ and $u+v=2y$.
Adding $(1)$ and $(2)$ gives,
$2\cdot5^c=2^u+2^v$
Since $v>u\geq0$, $v$ must be positive. So, $u\neq0$ since otherwise the left hand side would be even while the right hand side would be odd.
So, $u$ and $v$ both positive integers.
So, we have, $5^c=2^{u-1}(2^{v-u}+1)$
$\Rightarrow 2\nmid2^{u-1}(2^{v-u}+1)$
$\Rightarrow u=1, v>1$
Therefore, we have,
$5^c-2^{v-1}=1$
By Catalan's conjecture, this equation has no solution if $v-1\geq2$.
So, $v-1=1$
$\Rightarrow 5^c-2=1$
But since $c\geq2$, we must have $5^c-2\geq23$ which is a contradiction.
Therefore, there is no solution when $c\geq2$.

So, the only solution is $(x,y,z)=(2,2,2)$.

Posted: **Mon May 24, 2021 10:52 pm**

Anindya Biswas wrote: ↑
Mon May 24, 2021 10:22 pm
IDK how it can be done with Fermat's last Theorem, allow me to use Catalan's conjecture instead.

Let me state this theorem (it's proven, but the name stayed like this anyway!),
Catalan's conjecture : If $x,y,m,n$ are positive integers where $m\geq2, n\geq2$, and they satisfies \[x^m-y^n=1\] then $(x,y,m,n)=(3,2,2,3)$.

Now solution to our problem :
Let's start with taking $\pmod{3}$,
$3^x+4^y=5^z$
$\Rightarrow 1\equiv(-1)^z\pmod{3}$
$\Rightarrow z=2c$ for some positive integer $c$.

Let's take $\pmod{4}$,
$(-1)^x\equiv1\pmod{4}$
$\Rightarrow x=2a$ for some positive integer $a$.

Let's manipulate the equation,
$2^{2y}=5^{2c}-3^{2a}$

Case 1 : $c=1$.
In this case, we get $4^y=25-9^a$ and the only solution is, $a=1, y=2$. Because when $a\geq2, 25-9^a<0$.
So, this case gives us the solution $(x,y,z)=(2,2,2)$.

Case 2 : $c\geq2$.
$2^{2y}=(5^c-3^a)(5^c+3^a)$
Let's assume,
$5^c-3^a=2^u\cdots\cdots\cdots(1)$
$5^c+3^a=2^v\cdots\cdots\cdots(2)$
Where $u,v$ are nonnegative integers, $v>u$ and $u+v=2y$.
Adding $(1)$ and $(2)$ gives,
$2\cdot5^c=2^u+2^v$
Since $v>u\geq0$, $v$ must be positive. So, $u\neq0$ since otherwise the left hand side would be even while the right hand side would be odd.
So, $u$ and $v$ both positive integers.
So, we have, $5^c=2^{u-1}(2^{v-u}+1)$
$\Rightarrow 2\nmid2^{u-1}(2^{v-u}+1)$
$\Rightarrow u=1, v>1$
Therefore, we have,
$5^c-2^{v-1}=1$
By Catalan's conjecture, this equation has no solution if $v-1\geq2$.
So, $v-1=1$
$\Rightarrow 5^c-2=1$
But since $c\geq2$, we must have $5^c-2\geq23$ which is a contradiction.
Therefore, there is no solution when $c\geq2$.

So, the only solution is $(x,y,z)=(2,2,2)$.

Impressive.(By the way, don't know catalan

)(For using FLT you may try to prove $x=y=z$)
But however,It can be done in elementary way even more shorter than catalan's(slightly).
And your proof is very close to elementary one

maybe you should modify the last part a bit.

Posted: **Mon May 24, 2021 11:10 pm**

pythagorean triple?

Posted: **Tue May 25, 2021 7:58 am**

Dustan wrote: ↑
Mon May 24, 2021 11:10 pm
pythagorean triple?

Yeap.(But not so easy to prove).If I remember correctly this diophantine was solved by Sierpinski in 1956 and there is also a conjecture stating that if $a,b,c$ are phythagorean triples then $a^x+b^y=c^z$ has only the solution $(x,y,z)=(2,2,2)$. But sadly nobody could prove it

Posted: **Tue May 25, 2021 9:14 am**

Anindya Biswas wrote: ↑
Mon May 24, 2021 10:22 pm
$5^c-2^{v-1}=1$
By Catalan's conjecture, this equation has no solution if $v-1\geq2$.

This part can be done by LTE.
We have to show that if $m,n$ are positive integers satisfying $5^m-2^n=1$, then $(m, n)=(1,2)$.

Proof :
Taking mod $3$, we get
$(-1)^m-(-1)^n\equiv1\pmod3$
This only make sense when $m$ is odd, $n$ is even.
Again, notice that $2^n=5^m-1$
Since $4\mid 5-1, 2\nmid5, 2\nmid1$, By LTE, we get
$n=v_2(4)+v_2(m)$
Since $m$ is odd, $v_2(m)=0$
$\therefore n=2$.
$\therefore 5^m=5\Rightarrow m=1$
$\therefore (m, n)=(1,2)$ is the only valid solution.

Posted: **Tue May 25, 2021 4:43 pm**

Asif Hossain wrote: ↑
Tue May 25, 2021 7:58 am

Dustan wrote: ↑
Mon May 24, 2021 11:10 pm
pythagorean triple?
Yeap.(But not so easy to prove).If I remember correctly this diophantine was solved by Sierpinski in 1956 and there is also a conjecture stating that if $a,b,c$ are phythagorean triples then $a^x+b^y=c^z$ has only the solution $(x,y,z)=(2,2,2)$. But sadly nobody could prove it

you can check official soln, imo SL 1991

Posted: **Tue May 25, 2021 9:50 pm**

Dustan wrote: ↑
Tue May 25, 2021 4:43 pm

Asif Hossain wrote: ↑
Tue May 25, 2021 7:58 am

Dustan wrote: ↑
Mon May 24, 2021 11:10 pm
pythagorean triple?
Yeap.(But not so easy to prove).If I remember correctly this diophantine was solved by Sierpinski in 1956 and there is also a conjecture stating that if $a,b,c$ are phythagorean triples then $a^x+b^y=c^z$ has only the solution $(x,y,z)=(2,2,2)$. But sadly nobody could prove it
you can check official soln, imo SL 1991

IT WAS IN IMO SL. (Maybe then IMO was proposing famous or well known problems as the infamous 1988 p6 was collected from Euler's diary or notes)

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