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Find all pairs of numbers $(p,n)$ in which $p$ is a prime number, $n$ is a positive integer such that $p^n + 144$ is a perfect square.

Anh Phùng
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1 Answers1

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We can rewrite the situation like this:

where $m$ is an integer:

$p^n + 144 = m^2$

Now, we can rearrange the equation as follows:

$p^n = m^2 - 144$

And when we factor it from the right side:

$p^n = (m-12) (m+12)$

We know that prime numbers are whole numbers greater than 1 that cannot be exactly divided by any whole number other than itself and 1, so for $p^n$ to be a prime number, one of the factors on the right side must be 1, and the other $p^n$.

This gives us two cases:

  1. $m + 12 = 1$ and $m - 12 = p^n$
  2. $m + 12 = p^n$ and $m - 12 = 1$

So, let's solve these cases:

$1st$: $m + 12 = 1$

This implies $m = -11$

But this case has no solution for prime $p$ because when we say $m = -11$ :

$p^n = m - 12 = -11 -12 =-23$

And $-23 ≠ p^n$ for prime $p$ and positive $n$ since $p^n$ can't be negative.

So, we're looking at case 2: $m - 12 = 1$

This implies $m = 13$

Substitute $m = p^n - 12$:

$13 = p^n - 12$:

$p^n = 25$

This case gives us a solution for $p$ and $n$:

$p$ = 5, $n$ = 2:

So, the only solution is $(5, 2)$. The pair $(p, n) = (5, 2)$ satisfies the given conditions.

basilinnia
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  • oh, wrong. (p;n)=(3;4).... – Anh Phùng Nov 26 '23 at 12:19
  • The answer provided ignores a third case: $m+12=p^a; m-12=p^b$, whence $p^a-p^b=24$ or $p^b(p^{a-b}-1=24$. Since $p\mid 24$, it is the case that $p\in {2,3}$. Looking at cases, we find the values reported in the comment by Anh Phung – Keith Backman Nov 26 '23 at 17:38