2026/003) Solve in positive integers $a^2=b^4 + 16b + 1$

 b has to even because if b is odd then we have RHS is of the form 4k +2 and it cannot be a perfects squares

now next square after $b^4$ is $(b^2+1)^2$ and the given expression can be a square(not necessary it will be) if greater than or equal to $(b^2+1)^2$

$b^4 + 16b + 1 >= (b^2+1)^2$

or $b^4 + 16b + 1 >= b^4 + 2b^2 +1 $

or $16b >= 2b^2$ or $ b <= 8$

So we need to try even values of b less or equal to 8 giving $b= 2 =>  a^2 = 49 =>a = 7$ 

$b =4 => a^2 = 256 + 64 +1 = 321$ no solution

$b = 6=> a^2 = 1395$ no solution

$b = 8 => a = b^2+1 = 65$

Solution $(7,2)$ and $(65,8)$