[Guest]

N is a positive integer ≥ 2 , M is any positive integer while P is a prime number.

Determine all possible triplet(s) (M, N, P) that satisfy the equation:

P^N + 144 = M²

Edited October 7, 2009 by K Sengupta

[Guest]

i don't have an elegant solution but so far i have

(13,2,5), (15,4,3) and (20,8,2)

[Guest]

P^N = M² - 12² = (M - 12) * (M + 12)

A power of a prime has only same numbers (the prime itself) as factors (apart from 1).

So the condition above is only possible if M - 12 = 1 => M = 13.

[Guest]

methinks - I like your approach but power of prime also has factors that are multiples of that prime. i.e. 2^4=16 which also has factors 2,8 and 4

[Guest]

me_is_fob got them all. Here is my proof:

start with what methinks did and you have p^n = (m-12)(m+12) but now break p^n into p^(n/2 + i) and p^(n/2 - i) so

p^(n/2 + i) * p^(n/2 - i) = (m-12)(m+12)

if you let i be positive or negative then it doesn't matter which way you divide it so

p^(n/2 + i) = (m+12)

p^(n/2 - i) = (m-12)

now subtract and you get

p^(n/2 + i) - p^(n/2 - i) = 24

or

24 = p^(n/2) * (p^i - p^-i) = p^(n/2) * (p^i - 1/p^i) = p^(n/2) * (p^2i - 1)/(p^i) = p^(n/2 - i) * (p^2i - 1)

but we defined p^(n/2 - i) = (m-12)

so 24 = (m-12) (p^2i - 1)

or

24/(m-12) = (p^2i - 1)

now i could be xxx.5 if N is an odd number, but 2i will always be an integer. M must be >=13 (otherwise the left side is a neg number <-2) and i >0 (otherwise right side would be between -1 and 0) and M must be <=36 (otherwise left side is a fraction)

finally, M must be a value that results in a whole number on the left so the possible values of M are:

13,14,15,16,18,20,24,36

Now testing each of them we find that only 13,15, and 20 result in powers of a prime