Determine all possible quadruplets (p, q, r, s) of positive integers with p< = q< = r satisfying the equation p! + q! + r! = 2^s.
let the max power of 2 in p = k
then 2^k(a+2^x*b+2^y*c)=2^s where a,b,c are odd
but (a+..) in the bracets is odd if x not equal to 0 and y not equal to 0. so, x=0 for sure. so, q can take 2 values either p or (p+1). we also know 'a' should be factor of c as r>=p. so, a shud be 1 as a is odd. So, k can only take values 0 or 1 bcoz 2^k *a = p! and no odd factors of p other than 1.
case(1): q=p,k=0 then q=p=1,2^s = 2+c*2^y, that implies 2*(2^(s-1)-1)=c*2^y, c is odd so y=1. therefore c=2^(s-1)-1
As y=1, (power of 2 in r!) -(power of 2 in p!) = 1. so, r = 2,3 which implies that s=2,3
case(2): q=p,k=1 then q=p=2, 2^s=4+c*2^y, that implies 2^2*(2^(s-2)-1) =c*2^y, c is odd so y=2. there c=2^(s-2)-1
As y=2, s=4, c=3,r=4 is one of the solution
case(3):q=p+1,k=0 no solutions bcoz p=1,k=2 but x shud be 0 here x=1.
case(4):q=p+1,k=1, p=2,q=3, we get 2^2*(2^(s-2)-1)=c*2^y
y=2, so, c should be multiple of 3 as r>=q,r can take values 3,4,5 which yields {c=3,s=4} for r=4, {c =15, s= 6} for r=5. for r =3 power of 2 in r! - power of 2 in p! is not equal to 2(=y). so, no solutions for r=3 in this case
finally sols are(1,1,2,2),(1,1,3,3),(2,3,4,5) and (2,3,5,7)
Edited on August 30, 2007, 5:56 am
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