BDMO Primes!

Rewriting gives: $p^n=(m+12)(m-12)$ . Let $m+12=p^i$ and $m-12=p^j$ and so $i+j=n$ and $i>j$ .

Then we have:

$\begin{aligned} p^j(p^{i-j}-1) &=p^i-p^j \\ &=(m+12)-(m-12)\\ &=24=5^0(5^2-1) =2^3(2^2-1)=3^1(3^2-1).\\ \end{aligned}$

From here we have $3$ solutions: $m=13,15,20$ with sum $\fbox{48}$ .

Okay here we go

Rearranging the equation and simple factoring gives us

${ p }^{ n }+144={ m }^{ 2 }\\ { p }^{ n }={ m }^{ 2 }-144\\ { p }^{ n }=(m-12)(m+12)\\ \\$

Also, $p^n$ can be rewritten as ${ p }^{ x }({ p }^{ n-x })$

So,

${ p }^{ x }({ p }^{ n-x }) = (m-12)(m+12)$

Therefore, assuming that $p^x < ({ p }^{ n-x })$ ,

$p^x=(m-12)$

${ p }^{ n-x }=(m+12)$

We can now say that

$p^x+24=p^{n-x}$

So the difference between two powers of $p$ is $24$ . Now let us start testing with prime numbers, starting from 2:

${ 2 }^{ 0 }+24\neq { 2 }^{ y }\\ { 2 }^{ 1 }+24\neq { 2 }^{ y }\\ { 2 }^{ 2 }+24\neq { 2 }^{ y }\\ { 2 }^{ 3 }+24={ 2 }^{ y },\quad and\quad y=5$

We can see that further adding 24 to further powers of too will not result in a power of two as once you get to $2^5$ , the difference between two consecutive powers of two is greater that 24.

Now we try with the next prime: 3

${ 3 }^{ 0 }+24\neq { 3 }^{ y }\\ 3^{ 1 }+24=3^{ y },\quad where\quad y=3\\ { 3 }^{ 2 }+24\neq { 3 }^{ y }$

Once again we can see that continuing would be futile as the difference between $3^3$ and $3^4$ is greater than $24$

Okay moving on to 5

${ 5 }^{ 0 }+24=5^{ y },\quad and\quad y=2\\ 5^{ 1 }+24\neq 5^{ y }$

For the same reason as before, we can see that continuing with powers of 5 would be futile

Also, we cannot continue with any prime numbers after 5, because, as you can see, as the primes get bigger, the value of $n$ gets smaller. However, the smallest value of $n$ has already been reached (0). So continuing with 7 will not give us any solutions.

So now we can see that $p^n$ must be :

${ 5 }^{ 0 },{ 3 }^{ 1 },\quad or\quad { 2 }^{ 3 }$

And:

${ 5 }^{ 0 }+12=13\\ 3^{ 1 }+12=15\\ { 2 }^{ 3 }+12=20$

Adding these three values of $m$ , we get:

$15+13+20 = \boxed{48}$

Rearranging to $p^n = m^2 -144 = (m+12)(m-12)$

In order for LHS to be a prime power, both $(m+12)$ and $(m-12)$ must be, with $(m-12 \mid m+12)$ .

$\frac{m+12}{m-12} = 1 + \frac{24}{m-12}$

Therefore $(m-12) | 24$ .

This gives $m = 13, 14, 15, 16, 18, 20, 24, 36$ .

Although not the most efficient way, there are few enough cases to test one by one.

This is only valid when $m = 13, 15, 20: Sum = \boxed{48}$

I have tried to give a simplified solution: One may think at the initial that there is huge set of numbers so sum can be anything. But it’s not like that. p^n=(m+12)(m-12)=p^i*p^j (say) So, p^i-p^j=(m+12)-(m-12)=24. ( Just think 5^3-5^2=100. And so p,i,j can’t be huge numbers.)
So thinking with the initial prime nos 2, 3 and 5 and taking m>12 (As p^n is positive) we get- ->m=13 for which p=5 ,i=2,j=0 ->m=15 for which p=3,i=3, j=1 ->m=20 for which p=2,i=5,j=3 So sum of all possible m’s are=13+15+20=48