More primes and more primes!

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More primes and more primes!

Post Number:#1  Unread postby aritra barua » Wed Jul 05, 2017 7:55 pm

Find every triplet of positive integers ($m,p,y$) such that the equation:$2^m*p^2+1$=$y^7$ is satisfied when $p$ and $y$ are primes.
aritra barua
 
Posts: 45
Joined: Sun Dec 11, 2016 2:01 pm

Re: More primes and more primes!

Post Number:#2  Unread postby aritra barua » Fri Jul 07, 2017 10:45 am

Apparently,this is a TST problem from Hong Kong.We rewrite our equation:$2^m*p^2$=$y^7-1$=$(y-1)(y^6+y^5....+y+1)$.Clearly $y$ being odd implies $y^6+y^5...+y+1$ is odd.Hence $2^m$ must divide $y-1$ since $p$ is a positive integer.Now the equation $\frac{y-1}{2^m}$*$(y^6+y^5...+y+1)$=$p^2$ leads us to the straightforward solutions of $\frac{y-1}{2^m}$ and the latter one in the $L.H.S$ which are ($1,p^2$),($p^2,1$),($p,p$) since $p$ is a prime.But $\frac{y-1}{2^m}$ is strictly less than $y^6+y^5....+y+1$ which gives us the only solution ($1,p^2$).Hence,$y-1$=$2^m$ and gives us the equation $p^2$=$y^6+y^5....+y+1$=$\frac{y^7-1}{y-1}$=$\frac{(2^m+1)^7-1}{2^m}$=$(2^m)[(2^m+1)^6+(2^m+1)^5.... +(2^m+1)+1]$/$2^m$.$[(2^m+1)^6+(2^m+1)^5....+(2^m+1)+1]$ is congruent to $7$ (mod $2^m$) hence is equal to $2^m*a+7$ for some integers $a$.The equation can again be rewritten as $2^m*a$=$p^2-7$.If $m$ is greater than or equal to $2$,$4$|$p^2-7$ but the dividend is congruent to $2$ (mod $4$),thus gives us no solution.Hence $m$=$1$.The rest of the part is easy to be dealt with.
aritra barua
 
Posts: 45
Joined: Sun Dec 11, 2016 2:01 pm


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