There is method to my madness

Number Theory Level 5

Find the sum of all $p,q,s$ such that

$2^m.p^2+1=q^s$

and $p,q,s$ are odd primes and $m$ is not divisible by 4.

Clarification :

For every such triple, add the values of $p,q$ and $s$ to obtain the answer.

The answer is 19.

3 solutions

Souryajit Roy
Sep 19, 2015

$p|(q^{s}-1)$ .Note that $p$ cannot divide $q-1$ as it is even.

So, $p$ must divide $q^{s-1}+q^{s-2}+...1$ .

Now, $q^{s-1}+...+1$ is purely odd.

So we have, $2^{m}=q-1$ or, $q=2^{m}+1$

Now, $m=2^{t}$ So, $t=0$ or $1$

So, we have to equations-(a) $2p^{2}=3^{s}-1$ and (b) $4p^{2}=5^{s}-1$

Note that (b) has no solutions and (a) only has $(11,3,5)$ has solution.

u didn't explain why $p$ does not divide $q-1$ .

Here's the reason.

$(q-1)(q^{s-1} + q^{s-2} + .... +1) = p^{2} 2^{m}$ Since, $(q^{s-1} + q^{s-2} + .... +1)$ is odd and greater than $1$ , it is must and should be divisible by $p$ .

Then, either $p|q-1$ or $p^{2}|(q^{s-1} + q^{s-2} + .... +1)$ . The latter one can be solved easily, let's check the first one.

If $p|q-1$ , let us observe what happens.

Since, $q^{s-1} + q^{s-2} + .... +1$ is odd, it forces that $q^{s-1} + q^{s-2} + .... +1=p$ and $2^{m} | q-1$ . So, $q-1=2^{m}p$ .

So, $q=2^{m} p + 1$ . But $s>2$ , so, $q^{s} > q^{2} = 2^{2m}p^{2} + 2^{m+1}p + 1 > 2^{m}p^{2} +1$ , a contradiction. So, $p|q-1$ .

So, the latter one follows.

Surya Prakash - 5 years, 8 months ago

Bogdan Simeonov
Feb 28, 2015

Since the solution is very big, I willl divide it into two parts.

To start with, obviously $q^s \equiv 1\pmod{2^m}$ .Let v be the order of q modulo 2.Then v must divide s and $\varphi(2^m)=2^{m-1}$ .But s is odd, so v must be 1.So now we have that $q \equiv 1 \pmod{2^m}$ .We let $q=2^m.k+1$ .But since q-1 divides $q^s-1$ , we have that $2^m.k|2^m.p^2$ .That means that k is 1,p or $p^2$

Case 1:k=1.

In this case $q=2^m+1$ , so q is a Fermat prime.That means that m must be a power of 2.But m is not divisible by 4, so m is 1 or 2.This gives us two equations which we will handle in the next part, namely $2p^2+1=3^s$ and $4p^2+1=5^s$ .

Case 2:k=p

$2^m.p^2+1=(2^m.p+1)^s$ .

But expanding (using the Binomial theorem),we get that p must divide s, so p is equal to s.But then $2^{mp}.p^p>2^m.p^2$ and these are the first terms of the RHS and LHS respectively.So RHS>LHS, a contradiction.

Case 3:k=p^2

Here we would have that s=1, clearly not possible, since s is a prime number.

But that is just the first part, since we have yet to solve the equations from Case 1. But they can be solved using unique factorization in $Z(i)$ and $Z(\sqrt{-2})$ .

Niranjan Khanderia
Aug 17, 2018

$m=0~makes~ LHS~as~even.~But~RHS~is~ALWAYS~odd. \therefore~m\neq 0.\\ To~start~with~trial~~~ m=1,~2,~and~ p=1.\\ Find~ out~the~integer~we~get~from~\dfrac{\sqrt{q^s-1}}{2^m}~~where~q=3,~5,~7,~11,~~for~each~of~s=3,~5,~7. \\ If~we~get~an~integer~it~is~p~for~those~values~of~other~variables.\\ With ~in~ the~ rage~there~was~only~ONE~solution.~Only~integer~answer~was~11.\\ So~one~solution~is~(p,q,s)=(11,3,5)\\ Tried~other~bigger~primes,~and~m=3,5,6.~none~gave~an~integer.\\ So~p+q+s=11+3+5= \huge \color{#D61F06}{19}.\\$

Can any one give a simple proof that there is only one solution, or this is the only solution? Thanks.

There is method to my madness

The answer is 19.

3 solutions

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