first I will prove a little theorem that will be used to prove that (2,2) and (2,3) are the only solutions.
Theorem: if N is an integer greater than 2 then there exists a prime P less than N such that P is not a factor of N.
Proof: I will prove this by contradiction. Assume that N has as its prime factors every prime less than N. Name these primes p1,p2,...,pk. So then obviously N>=p1*p2*..*pk now let M=(p1*p2*...pk)-1 now obviously M is not divisible by any of the primes p1,p2,..pk so either M is itself prime or there exists a prime less than M that is not one of p1,p2,...pk. Either way this contradicts that p1,p2,...,pk is all the primes less than N and thus N can not have all primes less than N as its factors, with the obvious exception of N=2.
Having proven this I will first find all solutions with P=2
p!+q!=p^q then becomes 2+q!=2^q
now obviously q=1 does not work, but q=2 and q=3 do. Now to show there are no more solutions with q>3.
now with q>3 (q!/2) is still even and thus (1+(q!/2)) is odd, but 2^q can not have any odd divisors and thus 2+q!=2*(1+(q!/2)) can never equal 2^q for any q>3.
now to show no more solutions exist with p>2.
first let q>=p then we have p! as a factor of p!+q! and thus
p!+q!=p!*(1+(p!/q!)) and so every prime less than p divides this. But from my theorem we have that there exists a prime less than p that does not divide p and thus is not a factor of p^q thus p!+q! and p^q have at least 1 prime divisor that they do not share and thus can not be equal.
now for q<p. Let r be such that 2^r<=q!<2^(r+1) and let s be such that 2^s<=p<2^(s+1) then p!+q! has a factor of q! and thus a factor of 2^r and since p!/q! is even then (p!/q!)+1 is odd. So 2^r is the largest power of 2 that divides p!+q!. Similarly (2^s)^q=2^(s*q) is the largest power of 2 that divides p^q and for p^q=p!+q! then we have that r=q*s
but there is a well known process for finding r and that is with the summation of floor(q/2^k) for all k>0 now this is strictly less than the summation of q/2^k k>0 and that is equal to q thus r<q and thus the only possible integer solution to r=q*s is r=s=0 but if q>=2 then r can not be zero so the only possible solutions is if q=1 but this can not be a solution be because p!+1 is obviously larger than p^1.
Thus the only solutions are (2,2) and (2,3).
*edited to fix typo in second to last paragraph where I had (2^r)^q=2^(r*q) when I meant to have (2^s)^q=2^(s*q) *
Edited on December 26, 2008, 6:47 am
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Posted by Daniel
on 2008-12-22 00:41:08 |
solution with proof
