diophantine china

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 Joined: Thu Dec 09, 2010 3:30 pm
diophantine china
Find all the pairs of prime numbers $(p,q) $such that $pq5^p+5^q.$
Re: diophantine china
It is trivial to note that $(5,2), (5,5)$ is two of the solution(s). I guess these are the only ones.
Lets consider, $p,q \neq 5$ and $p>q$
Then we arrive at the condition: $pq\left (5^{pq} + 1 \right)$
and hence $p\left (5^{pq} + 1 \right)$ , $q\left (5^{pq} + 1 \right)$
Applying Fermat's Little theorem along with the condition $5^{pq} \equiv 1 (mod \ p,q)$, we obtain
$p5^p  5, 5^q + 5$
$q5^q  5, 5^p + 5$
Hence, $pq \left (5^p + 5 \right) \left (5^q + 5 \right)$
$\Rightarrow pq5^{p+q} + 5^2$
Using the condition $pq5^{pq} + 1$ we obtain now
$pq\left (5^{p+q} + 5^2 \right)\left (5^{pq} + 1 \right)$
$\Rightarrow pq\left( 5^{2p} +5^{p+q} + 5^{p+2q} + 5^2 \right)$
$\Rightarrow pq\left(5^{2p} + 5^{p+2q} \right)$
$\Rightarrow pq\left(5^p + 5^{2q} \right)$
This gives us the condition $q = 2$
However, easy scratching reveals that using this condition we are left with no acceptable value of $p$. That closes it.
Choosing $p=5$ and a distinct $q$, we can show by similar arguments that $q=2$
Let me know if I'm wrong
Lets consider, $p,q \neq 5$ and $p>q$
Then we arrive at the condition: $pq\left (5^{pq} + 1 \right)$
and hence $p\left (5^{pq} + 1 \right)$ , $q\left (5^{pq} + 1 \right)$
Applying Fermat's Little theorem along with the condition $5^{pq} \equiv 1 (mod \ p,q)$, we obtain
$p5^p  5, 5^q + 5$
$q5^q  5, 5^p + 5$
Hence, $pq \left (5^p + 5 \right) \left (5^q + 5 \right)$
$\Rightarrow pq5^{p+q} + 5^2$
Using the condition $pq5^{pq} + 1$ we obtain now
$pq\left (5^{p+q} + 5^2 \right)\left (5^{pq} + 1 \right)$
$\Rightarrow pq\left( 5^{2p} +5^{p+q} + 5^{p+2q} + 5^2 \right)$
$\Rightarrow pq\left(5^{2p} + 5^{p+2q} \right)$
$\Rightarrow pq\left(5^p + 5^{2q} \right)$
This gives us the condition $q = 2$
However, easy scratching reveals that using this condition we are left with no acceptable value of $p$. That closes it.
Choosing $p=5$ and a distinct $q$, we can show by similar arguments that $q=2$
Let me know if I'm wrong
"Je le vois, mais je ne le crois pas!"  Georg Ferdinand Ludwig Philipp Cantor
Re: diophantine china
What about $(3,2)$?Avik Roy wrote:It is trivial to note that $(5,2), (5,5)$ is two of the solution(s). I guess these are the only ones.
Lets consider, $p,q \neq 5$ and $p>q$
Then we arrive at the condition: $pq\left (5^{pq} + 1 \right)$
and hence $p\left (5^{pq} + 1 \right)$ , $q\left (5^{pq} + 1 \right)$
Applying Fermat's Little theorem along with the condition $5^{pq} \equiv 1 (mod \ p,q)$, we obtain
$p5^p  5, 5^q + 5$
$q5^q  5, 5^p + 5$
Hence, $pq \left (5^p + 5 \right) \left (5^q + 5 \right)$
$\Rightarrow pq5^{p+q} + 5^2$
Using the condition $pq5^{pq} + 1$ we obtain now
$pq\left (5^{p+q} + 5^2 \right)\left (5^{pq} + 1 \right)$
$\Rightarrow pq\left( 5^{2p} +5^{p+q} + 5^{p+2q} + 5^2 \right)$
$\Rightarrow pq\left(5^{2p} + 5^{p+2q} \right)$
$\Rightarrow pq\left(5^p + 5^{2q} \right)$
This gives us the condition $q = 2$
However, easy scratching reveals that using this condition we are left with no acceptable value of $p$. That closes it.
Choosing $p=5$ and a distinct $q$, we can show by similar arguments that $q=2$
Let me know if I'm wrong
One one thing is neutral in the universe, that is $0$.
Re: diophantine china
I think you deduced this from $2q\ge 0,$ but note $p5^{2q}+5^p\rightarrow p5+5^{2q}=>p5^{q1}+1$ which we found true.
Ishfaq,was it a problem from China?
Ishfaq,was it a problem from China?
One one thing is neutral in the universe, that is $0$.
Re: diophantine china
Sorry for this statement. $(3,2)$ is obtained by letting $q=2$ and observing that $2p5^p + 1$ and $p5^p 5$Avik Roy wrote: However, easy scratching reveals that using this condition we are left with no acceptable value of $p$. That closes it.
Masum, I didn't get your argument. However, my logic was simpler I guess. How can we expect a prime dividing something fractional
"Je le vois, mais je ne le crois pas!"  Georg Ferdinand Ludwig Philipp Cantor
Re: diophantine china
this prime does not divide the fraction,only the numerator with denominator coprime to $p$
What I meant is $p5+5^{2q}=5(5^{1q}+1)\rightarrow p\frac {5^q+5} {5^q}$ and since $gcd(5,p)=1,p5^{q1}+1$ which we previously found.So I doubted.Did you really deduce this from $2q\ge 0$,I didn't understand this.If so then $p5(5^{1q}+1)=> 1=q$,a contradiction.
What I meant is $p5+5^{2q}=5(5^{1q}+1)\rightarrow p\frac {5^q+5} {5^q}$ and since $gcd(5,p)=1,p5^{q1}+1$ which we previously found.So I doubted.Did you really deduce this from $2q\ge 0$,I didn't understand this.If so then $p5(5^{1q}+1)=> 1=q$,a contradiction.
One one thing is neutral in the universe, that is $0$.