Let $d=(a,b)$ with $a=da_1$ and $b=db_1$. We then get $a_1-b_1 = d(5b_1^2-4a_1^2)$. I now claim $5b_1^2 - 4a_1^2=1$. Assume the contrary, let $p\mid 5b_1^2-4a_1^2$ be a prime. Then $p\mid a_1-b_1$. So, $5b_1^2-4a_1^2\equiv a_1^2\pmod{p}$, forcing $p\mid a_1,b_1$, a contradiction. Having established this, we get \[ a_1 - b_1 = d \quad\text{and}\quad 5b_1^2 - 4a_1^2=1. \]Plugging these values in, we get immediately \[ 8b+1 = 8b_1d+1 = (b_1-2d)(b_1+2d). \]If $b_1-2d>1$ this is not prime. If $b_1=2d+1$ then $8b+1 = 8d(2d+1)+1 = (4d+1)^2$, which is not a prime. Remark. Very similar problem appeared in Iran 1997 and France TST 2005.

Well-known lemma: if $a,b,c,d$ are four natural numbers and $ab=cd$, then $a+b+c+d$ is composite. Let $c=a-b \in \mathbb{N}$. Then $a=b+c$ and condition is equivalent to $c=b^2-8bc-4c^2$ which can be rewritten how $c(4c+1)=b(b-8c)$. Now And now multiply both parts of the equality by 12! We get $(12c)(4c+1)=(6b)(2b-16c)$, so by lemma $12c+(4c+1)+6b+(2b-16c)=8b+1$ is composite and we are done!

ignore wrong sol

Let \( p = 8b + 1 \). Then we have: \[ b^2 \equiv_p (a - b)(4a + 4b + 1) \\ \implies 2b^2 \equiv_p (a - b)(8a + 1) \iff 8a^2 - 8a b + a - b \equiv_p 8a^2 + 2a - b \equiv_p 2b^2 \\ \iff 2(2a - b)(2a + b) + 2a - b = (2a - b)(4a + 2b + 1) \equiv_p 0 . \] Hence if \( p \) is a prime, then \( p \mid 2a - b \) or \( 4a + 2b + 1 \). However, since \( a(4a + 1) = b(5b + 1) \) as \( a > b \), it follows that \( 4a + 1 < 5b + 1 \iff a < \frac{5}{4}b \), so \( 2a - b < \frac{3}{2}b < p \) and \( 4a + 2b + 1 < 7b + 1 < p \), a contradiction.