Need to confirm if this is right

For discussing Olympiad Level Number Theory problems
Asif Hossain
Posts: 171
Joined: Sat Jan 02, 2021 9:28 pm

Need to confirm if this is right

Unread post by Asif Hossain » Thu Apr 22, 2021 12:56 pm

Problem: Find all positive integers $x,y$ such that $p^x-y^p=1$ where $p$ is prime.(Czech Slovakia 1996)
Solution:(Need to confirm :oops:)
We can rewrite this as $p^x=1+y^p$
Now $p|1+y$ otherwise $1+y=1$ or $y=0$ which is not positive integer
So, Applying LTE we can get $\nu_{p}(1+y)=x-1$. Since the LHS is $p^x$ it must happen that $1+y=p^{x-1}...(1)$
Now we can write the question as $p^x -y^p=p^{x-1}-y \Rightarrow p^{x-1}(p-1)=y(y^{p-1}-1) \Rightarrow (1+y)(p-1)=y(y^{p-1}-1) \Rightarrow y|p-1$
but from $(1)$ we already have $p-1|y$ so $y=p-1$
So now the question downs to $p^x-(p-1)^{p}=1$
Now again applying LTE it is easy to see $x=2 \Rightarrow p^2-(p-1)^p=1$ which implies $p=3$ yieldin $(p,x,y)=(3,2,2)$. So the left case is only $p=2$ which yields $(p,x,y)=(2,1,1)$ and these are the only solution.
$\square$
Hmm..Hammer...Treat everything as nail

~Aurn0b~
Posts: 43
Joined: Thu Dec 03, 2020 8:30 pm

Re: Need to confirm if this is right

Unread post by ~Aurn0b~ » Thu Apr 22, 2021 2:07 pm

Asif Hossain wrote:
Thu Apr 22, 2021 12:56 pm
Problem: Find all positive integers $x,y$ such that $p^x-y^p=1$ where $p$ is prime.(Czech Slovakia 1996)
Solution:(Need to confirm :oops:)
We can rewrite this as $p^x=1+y^p$
Now $p|1+y$ otherwise $1+y=1$ or $y=0$ which is not positive integer
So, Applying LTE we can get $\nu_{p}(1+y)=x-1$. Since the LHS is $p^x$ it must happen that $1+y=p^{x-1}...(1)$
Now we can write the question as $p^x -y^p=p^{x-1}-y \Rightarrow p^{x-1}(p-1)=y(y^{p-1}-1) \Rightarrow (1+y)(p-1)=y(y^{p-1}-1) \Rightarrow y|p-1$
but from $(1)$ we already have $p-1|y$ so $y=p-1$
So now the question downs to $p^x-(p-1)^{p}=1$
Now again applying LTE it is easy to see $x=2 \Rightarrow p^2-(p-1)^p=1$ which implies $p=3$ yieldin $(p,x,y)=(3,2,2)$. So the left case is only $p=2$ which yields $(p,x,y)=(2,1,1)$ and these are the only solution.
$\square$
I think its correct, however, the solution could be more short, after figuring out, $1+y=p^{x-1}$(Assuming p is odd ofc) , we can say that $\frac{y^p+1}{y+1}=p\Rightarrow 1+y \geq \frac{y^p+1}{y+1} \Rightarrow y^2+2y\geq y^p \Rightarrow y+2\geq y^{p-1}\Rightarrow 2\geq y(y^{p-2}-1)$

Asif Hossain
Posts: 171
Joined: Sat Jan 02, 2021 9:28 pm

Re: Need to confirm if this is right

Unread post by Asif Hossain » Thu Apr 22, 2021 2:15 pm

Nice Solution :o the idea of bounding didn't come to my mind. :cry: mine was rather unnecessarily long :(
Hmm..Hammer...Treat everything as nail

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