MHB P is the sum of 2 consecutive squares

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An odd prime \( p \) can be expressed as the sum of two consecutive squares, \( p = a^2 + (a+1)^2 \), if and only if it takes the form \( p = \frac{u^2 + 1}{2} \). The derivation shows that \( p = 2a^2 + 2a + 1 \) leads to \( 2p = (2a + 1)^2 + 1 \). This formulation holds true regardless of whether \( p \) is prime, as demonstrated with the example of \( a = 6 \), yielding \( p = 85 \), which is not prime. The conclusion is that the relationship between odd primes and sums of consecutive squares is valid under specific conditions, but the prime status of \( p \) is not a requirement. The discussion confirms the mathematical validity of the derived expressions.
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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 $p=\dfrac{u^2+1}{2}$.
 
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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.
 
Alexmahone said:
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
 
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