## diophantine china

For discussing Olympiad Level Number Theory problems
ishfaqhaque
Posts: 20
Joined: Thu Dec 09, 2010 3:30 pm

### diophantine china

Find all the pairs of prime numbers $(p,q)$such that $pq|5^p+5^q.$

Avik Roy
Posts: 156
Joined: Tue Dec 07, 2010 2:07 am

### 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: $pq|\left (5^{p-q} + 1 \right)$
and hence $p|\left (5^{p-q} + 1 \right)$ , $q|\left (5^{p-q} + 1 \right)$
Applying Fermat's Little theorem along with the condition $5^{p-q} \equiv -1 (mod \ p,q)$, we obtain-
$p|5^p - 5, 5^q + 5$
$q|5^q - 5, 5^p + 5$
Hence, $pq| \left (5^p + 5 \right) \left (5^q + 5 \right)$
$\Rightarrow pq|5^{p+q} + 5^2$
Using the condition $pq|5^{p-q} + 1$ we obtain now
$pq|\left (5^{p+q} + 5^2 \right)\left (5^{p-q} + 1 \right)$
$\Rightarrow pq|\left( 5^{2p} +5^{p+q} + 5^{p+2-q} + 5^2 \right)$
$\Rightarrow pq|\left(5^{2p} + 5^{p+2-q} \right)$
$\Rightarrow pq|\left(5^p + 5^{2-q} \right)$
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

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: diophantine china

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: $pq|\left (5^{p-q} + 1 \right)$
and hence $p|\left (5^{p-q} + 1 \right)$ , $q|\left (5^{p-q} + 1 \right)$
Applying Fermat's Little theorem along with the condition $5^{p-q} \equiv -1 (mod \ p,q)$, we obtain-
$p|5^p - 5, 5^q + 5$
$q|5^q - 5, 5^p + 5$
Hence, $pq| \left (5^p + 5 \right) \left (5^q + 5 \right)$
$\Rightarrow pq|5^{p+q} + 5^2$
Using the condition $pq|5^{p-q} + 1$ we obtain now
$pq|\left (5^{p+q} + 5^2 \right)\left (5^{p-q} + 1 \right)$
$\Rightarrow pq|\left( 5^{2p} +5^{p+q} + 5^{p+2-q} + 5^2 \right)$
$\Rightarrow pq|\left(5^{2p} + 5^{p+2-q} \right)$
$\Rightarrow pq|\left(5^p + 5^{2-q} \right)$
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
What about $(3,2)$?
One one thing is neutral in the universe, that is $0$.

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: diophantine china

I think you deduced this from $2-q\ge 0,$ but note $p|5^{2-q}+5^p\rightarrow p|5+5^{2-q}=>p|5^{q-1}+1$ which we found true.
Ishfaq,was it a problem from China?
One one thing is neutral in the universe, that is $0$.

Avik Roy
Posts: 156
Joined: Tue Dec 07, 2010 2:07 am

### Re: diophantine china

Avik Roy wrote: However, easy scratching reveals that using this condition we are left with no acceptable value of $p$. That closes it.
Sorry for this statement. $(3,2)$ is obtained by letting $q=2$ and observing that $2p|5^p + 1$ and $p|5^p -5$

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

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
this prime does not divide the fraction,only the numerator with denominator co-prime to $p$
What I meant is $p|5+5^{2-q}=5(5^{1-q}+1)\rightarrow p|\frac {5^q+5} {5^q}$ and since $gcd(5,p)=1,p|5^{q-1}+1$ which we previously found.So I doubted.Did you really deduce this from $2-q\ge 0$,I didn't understand this.If so then $p|5(5^{1-q}+1)=> 1=q$,a contradiction.
One one thing is neutral in the universe, that is $0$.