WakeUp wrote: Determine all positive integers $n$ for which $2^{n+1}-n^2$ is a prime number. .. $ n >3$ If $ n$ is odd then ${n+1}$ is even $ \Rightarrow$w $2^{n+1}-n^2$ is not a prime number If $n$ is even then Let $2^{n+1}-n^2 =p $ ($p$ is a prime number) because $2^{n+1}-n^2 \equiv 0 (mod 4) $ $\Rightarrow p$ is not a prime number So $n=1;3$

magical wrote: WakeUp wrote: Determine all positive integers $n$ for which $2^{n+1}-n^2$ is a prime number. If $ n$ is odd then ${n+1}$ is even $ \Rightarrow 2^{n+1}-n^2$ is not a prime number If $n$ is even then Let $2^{n+1}-n^2 =p $ ($p$ is a prime number) $\Leftrightarrow 2^{n+1}=n^2 +p \equiv 0 ( mod 4) $ because $p=4k+1;4k+3$ and $ n \equiv 0 (mod 4)$ .... Don't understand your argument with $\pmod{4}$.. If $n=2k$ is even then $p=2^{n+1}-n^2=2^{2k+1}-4k^2$ forcing $p$ to be even, i.e. $p=2$. Then dividing by $2$ we have $2^{2k}=2k^2+1$ which by inequalities or by reasoning with parity, can only happen when $k=0$. But then $n=0$, but $n>0$, so no solutions. Otherwise we assume $n=2k+1$ where $k\ge 0$ to obtain the equation $(2^{k+1}-2k-1)(2^{k+1}+2k+1)=p$. Since $p$ is prime we must have $2^{k+1}-2k-1=1$ and $2^{k+1}+2k+1$ is a prime. When $k=0,1$ it works but it won't otherwise since $2^{k+1}-2k-1>1$ for $k>1$. This corresponds to the solutions $n=1,3$. Hence the sole solutions $n=1,3$. I never understand why so many problems tighten to positive when not necessary, in this case $n=0$ is left out.

magical wrote: WakeUp wrote: Determine all positive integers $n$ for which $2^{n+1}-n^2$ is a prime number. .. $ n >3$ If $ n$ is odd then ${n+1}$ is even $ \Rightarrow$w $2^{n+1}-n^2$ is not a prime number Please. Why?

I am explaining.Let $n=2k-1$.Then $2^{n+1}-n^2=(2^k+n)(2^k-n)$,so composite except $2^k-2k+1=1$ or $k=1,n=1$

I made a silly mistake.Here $2^{k-1}=k$ means $k=1,2;n=1,3$.The case $n=0$ yields $2$,and it is divisible by $4$when $n$ even,so not prime .