Find four positive integers p,q,r and s such that:
- p*q*r*s= 8!, and:
- p*q+p+q = 524, and:
- q*r+q+r = 146, and:
- r*s+r+s = 104
Prove that there are no others.
By the way, if negative integers had been allowed, there would still be only one solution
a) p*q*r*s = 8!, so the only prime factors in these numbers are 7 and less.
b) Subtracting equation 4 from 3 and rearranging terms gives
(r+1)(q-s) = 42,
so if r is negative then (r+1) must be -1,-2,-3,-6,-7,-14,-21 or -42
This makes r = -2,-3,-4,-7,-8,-15,-22 or -43.
We can rule out -22 and -43 (prime factors are too large),
so r = -2,-3,-4,-7,-8, or -15
c) From equation 3, q = (146-r)/(1+r)
The possibilities are
r q
-- --
-2 -148 <== multiple of 37
-3 -74.5 <== not an integer
-4 -50 <== multiple of 25
-7 -25.5 <== not an integer
-8 -22 <== multiple of 11
-15 -11.5 <== not an integer
so, r must be positive, and the only solution is the one previously posted
Negative solutions
