2022/059) If $4x=a^2+3$ where (a,x) are positive integers & x is not a multiple of 3, then It's only possible when x are certain positive integers of the form of 6y+1. Prove this?

 Because 4x is even so we have a is odd 

So let a = 2n + 1

$4x= (2n+1)^3 + 3 = 4n^2 + 4n + 4$

or $x= n^2 + n + 1 = n(n+1) + 1$

now n can be 0/1/2 mod 3

if n is one of the form $(3k, 3k+1, 3k-1$

If n is of the form 3k then $x-1 = 3k(3k+1)$ divisible by 6 so x of the orm 6m + 1

if n is of the form 3k + 1 then $x-1 = (3k+ 1)(3k+2) = 9k^2 + 9k^2 + 2 = 9k(k+1) + 2$ remainder 2 when devided by 6 

so x = 6m + 3 but is divisible by 3 so not admissible

if n is 3k -1 then 

$x-1 = (3k-1)(3k)$ of the form 6m or x+ 1of the form 6m + 1

So it is of the form 6m + 1(or 6y + 1)  and hence proved