.

[rain estuary]

Is p prime?

I'll assume so

Consider $m^2-60^2=p^n$. You can factorize left hand side to get $(m+60)(m-60)=p^n$. Now, invoke unique factorization theorem. We deduce that $m+60=p^i$ and $m-60=p^j$ where $i+j=n$ and $j<i$.

Consider $p^i-p^j=p^j(p^{i-j}-1)=120=2^3\times 3\times 5$. Therefore, $p^j$ has three possible choices:

If $p=3$, then $j=1$, but then $3^{i-1}-1=40$. Impossible.

If $p=5$, then $j=1$, then $5^{i-1}-1=24$, so $i=3$, so we have our first solution $(p,m,n)=(5,65,4)$

If $p=2$, then $p^{i-j}-1$ has to be odd (since $j<i$) so $j=3$ and $2^{i-3}-1=15$, so $i=7$, which implies that $n=10$ and $m=2^3+60=68$, so we have our second solution $(p,m,n)=(2,68,10)$

[inland saffron]

Kevin S

[rain estuary]

In conclusion, there are only two triples $$(p,m,n)=(2,68,10) \text{ or } (5,65,4)$$

[inland saffron]

Kevin S

[rain estuary]

@minor marsh

[rain estuary]

What is your triple for p=11 then?

Thanks, I missed one case, which is that $p^j=1$, i.e., $j=0$ and then $p^{i}-1=120$, which implies that $p=11$ and then that gives the third one $(11,61,2)$.

[inland saffron]

Kevin S

[rain estuary]

unique factorization theorem essentially says that every natural number can be written as a product of primes in a unique way. This is very useful for this type of number theory problem. There might be tricky ways out there but unique factorization theorem usually gives the most straightforward solution.

that's quite impressive given your young age!

Just a general remark. The significance of unique factorization theorem (UFT) is that it shows that the set of natural numbers has a nice multiplicative structure, so if you can transform an equation (like the one in your question) from an additive one to a multiplicative one, then things can get much better.

by comparing sizes you can conclude that $a>b$ right?

[inland saffron]

Kevin S

[rain estuary]

so you can factorize, $2^b(2^{a-b}-1)=120=2^3\times 3\times 5\times 1$

[inland saffron]

Kevin S

[rain estuary]

now you need to observe the parity. On the left hand side, you have an even number times an odd number, on the right hand side, the only even part is $2^3$. This implies (you might need to think about it) that $2^b$ has to equal to $2^3$. (If not, then there exists an even factor that divides $2^{a-b}-1$ which is impossible because $a>b$)

[inland saffron]

Kevin S

[rain estuary]

No m should be 2^7-60.

Those essentially follow from unique factorization theorem. The idea is that 5 is the highest power of 5 dividing 120 (RHS), and this should be the same for left hand side (if let's say $5^2$ divides $p^j(p^i-1)$, then $5^2$ divides $p^j$ which implies $j\ge 2$, but then that means 25 divides 120, which is absurd) So, we have to have $j=1$ in these cases.

[inland saffron]

Kevin S

[rain estuary]

This one also. But you can also understand it from the perspective of parity. Consider $2^j(2^{i-j}-1)=2^3\times 3\times 5$. The left hand side is a product of an even number and an odd number. Powers of $2$ on the right hand side cannot divide the odd part, so $2^3$ must divide $2^j$. This shows that $j\ge 3$, but in fact $j=3$. (Suppose $j\ge 4$, then $2^4$ divides the left hand side, which implies $2^4$ divides 120, which is impossible)

[inland saffron]

Kevin S

[rain estuary]

ur welcome