Find values for which the given expression is a perfect square

Question

Find all prime numbers $p$ such that $38 p+23$ is a perfect square. $p$ can be $ 7, 11, 79$ etc. I think there would be infinitely many primes. Is there any method to determine all the solutions?

Thanks in advance.

Answer 1

Partial result:

From $$38p+19 = (x-2)(x+2)\implies x=19y\pm 2$$

we get $$2p+1= 19y^2\pm 4y \implies y=2z+1$$

so $$p =38(z^2+z)+9\pm 2(2z+1)$$

so $(A)\;\;p= 38z^2+42z+11\;\;\;\;$ or $\;\;\;\;\;(B):p=38z^2+34z+7$.

So $p$ is of that form. Now the story doesn't end...

Answer 2

Suppose :

$$38 p + 23 = (p +k)^2$$

Then we must have:

$$p^2+p(2k-38)+k^2-23=0$$

$$\Delta=-152 k +1536; k ≈ 10.1..$$

S0 $k ≤ 10$

This gives only primes 7 and 11.

Now suppose $38k+23=(p-k)^2$; then:

$$p^2-(2k+38)p+k^2-23=0$$

$$\Delta=152 k +1536 ≥ 0$$

$$k≤ -10 $$

which for example gives $-22$ for $p=79$ .

Since the numbers below $ -10$is infinite the number of solutions can be infinite.