P is the sum of 2 consecutive squares

[alexmahone]

Let $p$ be an odd prime.

Prove that $p$ is the sum of 2 consecutive squares i.e. $p=a^2+(a+1)^2$ if and only if $p$ has the form $p=\dfrac{u^2+1}{2}$.

 

[alexmahone]

I think I got it.

$p=a^2+(a+1)^2=2a^2+2a+1$

$2p=4a^2+4a+2=(2a+1)^2+1$

$p=\dfrac{(2a+1)^2+1}{2}=\dfrac{u^2+1}{2}$ where $u=2a+1$

Could someone confirm? I didn't use the fact that $p$ is prime.

 

[kaliprasad]

I think I got it.

$p=a^2+(a+1)^2=2a^2+2a+1$

$2p=4a^2+4a+2=(2a+1)^2+1$

$p=\dfrac{(2a+1)^2+1}{2}=\dfrac{u^2+1}{2}$ where $u=2a+1$

Could someone confirm? I didn't use the fact that $p$ is prime.

above is true whether p is prime or not as you proved it

as an example a= 6

$p = a^2 + (a+1)^2 = 6^2+7^2 = 85 = \dfrac{1}{2}(13^2+1)$ and 85 is not prime