Find all nonnegative integers $k, n$ which satisfy $2^{2k+1} + 9\cdot 2^k + 5 = n^2.$
Problem
Source: India Postal Set 3 P 5 2016
Tags: number theory, Diophantine equation
rafayaashary1
18.01.2017 06:54
So we have $2^k\mid n^2-5$. However by $\pmod{8}$ it follows that $2^k\in\{2,4\}$. Now we try cases and find only $\boxed{(k,n)=(0,4)}$
Ankoganit
18.01.2017 06:59
@above you missed $2^k=1$.
rafayaashary1
18.01.2017 07:01
Ankoganit wrote: @above you missed $2^k=1$. Oops I read "positive" instead of "nonnegative"
Djaliloo
18.01.2017 09:57
Assume $n > 2$. Working modulu 8 we get $n^{2} =5 (mod 8)$ which is impossible. So the only solution is $(k,n) = (0,4)$.
SHREYAS333
20.05.2019 23:13
Djaliloo wrote: Assume $n > 2$. Working modulu 8 we get $n^{2} =5 (mod 8)$ which is impossible. So the only solution is $(k,n) = (0,4)$. It should be $k > 2 $
Math-wiz
31.10.2019 08:30
Albania BMO TST 2014