A number theory problem by Bulbuul Dev
How many unordered triples of positive integers are there, such that for a positive integer , we have
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1 solution
Assume all p,q and r are >= 3 that would mean that 3 divides RHS absurd. so one of them is <= 2. WLOG let it be r i.e r = 2. The equation then becomes p! + q! = 2^s - 2 Again if p and q are both >= 4 RHS has to be divisible by 4 which is impossible so p or q is <= 3 i.e (Wlog) q = 2 or 3. q = 3 => p! = 2^s - 2 - 6 = 2^s - 8 if p >= 6 equality fails so p <= 5 testing we get p = 5,4 as solutions q = 2 => p! = 2^s - 4 if p >= 4 last equality fails so p <= 3, testing we get no further solutions conclusion : (2,3,4,5), (2,3,5,7)