A number theory problem by Daniel Remo

If n is even:

n=2k, so x^n-1=(x^k+1)(x^k-1)=p^r So p^r has factors with a difference of two, so clearly they are 2 and 4, and 3^2-1=2^3 is the only solution here.

If n is odd and x is even:

n>1 so x^n-1=7 (mod8) So p^r must also equal 7 (mod8) However, no prime satisfies this with r>1 So there are no solutions with n odd and x even.

If n is odd and x is odd:

p is even so it must equal 2 Now x^n-1=2^r So x^n=1 (mod8) However this condition cannot be satisfied with x and n odd (this is easy to show with some algebra and mods) So there are no solutions here either.

So the only solution is 3^2-1=2^3, giving an answer of 10.

I would suggest trying numbers. You're looking for 2 perfect powers that have a difference of 1. 8 and 9 come to mind, and sure enough, they fit the equation. $3^{2}-1=2^{3}$ , thus $2+3+2+3=10$ .

First you show that there is only one solution if n is even (difference of 2 squares, it is very 'hard' for 2 numbers to both be perfect powers of the same prime). Since perfect powers are 2 apart, and they are both divisible by p, this p must be 2. Now find the powers of 2 that are 2 apart, prove that they are the only ones, then multiply to get the full solution. Now we show that there are no solutions in odd numbers. Clearly, by an inductive idea, we need only to prove it when n is prime (through factorizing). Now suppose that n and p are not the same prime. Therefore they have common divisor of 1. Now using some LTE (lifting the exponent) stuff, a well-known lemma, that up(x^n-y^n)=up(x-y), where up is the largest positive integer n such that p^n divides a in up(a), and when p divides x-y, but not x and not y, and the gcd of n and p = 1. Setting y=1, we find that when p and n are separate primes, since both x^n-1^n and x-1 are both powers of p with the same power, they are equal, so x^n-1^n divided by x-1=1+x+x^2...+x^n-1=1, so x=0, which cannot happen. Now we consider the case when n=p, they are the same prime. We first deal with the case of p being an odd prime. The theorem of LTE states that if p is an odd prime, p divides x-y, but not x or y, then up(x^n-y^n)=up(x-y)+up(n). It does not specify any gcd criteria. So, since our new equation has become x^p-1^p=p^r, we find that, using the definition of up, and setting y=1, we get up(x^n-y^n)=up(x-1)+1, so through division, we get that up(1+x+x^2...+x^p-1)=1, ie 1+x+x^2...+x^p-1=p. But x-1 is at least 1, or p. x-1 is clearly not greater than or equal to 1+x+x^2...+x^p-1, so x-1=1, so x=2. Contradiction. Now we deal with the case of x=2. If 2^p-1=p^r. By Fermat's Little Theorem, 2^p-2 is congruent to 0 mod p, but by are hypothesis, it is also congruent to 1 mod p as well (2^p-1=p^r, and is therefore congruent to 0 mod p), which is a contradiction, since p>1. So the only solutions are n=2, x=3, p=2, r=3. Comment if you have problems.