(a, b, c) = (2,4,8);(3, 5, 15).
EXPLANATION:
We substitute (a-1, b-1, c-1) =(p, q, r)
By the problem, (p+1)(q+1)(r+1) -1 is a multiple of pqr.
Or, (p+1)(q+1)(r+1) - 1 = pqrs, where s is an integer > =2 ....(i)
Or, 1/p + 1/q + 1/r + 1/(pq) + 1/(pr) + 1/(qr) = s-1....(*)
Now, 1< a< b< c gives p> = 1, q> = 2, r> =3
Accordingly, from (*), we obtain:
s-1 < = 1+ ½ + 1/3 + ½ + 1/3 + 1/6 < 3
Or, s< 4
Hence s = 2, 3
If s=2, then 1/p< 1, giving p> 1, so that p> =2
But, if p> =3, then q> =4, r> =5
Accordingly,
(s-1) < = 1/3 + ¼+ 1/5 + 1/12 + 1/15 + 1/20 = 59/60 < 1
Thus p=2, whenever s=2
Substituting this in (i) and simplifying, we obtain:
(q-3)(r-3) = 11, giving (q, r) = (4, 14) subject to the constraint q< r.
Thus, (a, b, c) = (3, 5, 15)
For s=3, we have 1/p< 2, so that: p> ½, giving: p> =1
If p> = 2, then (s-1) < = ½+ 1/3 + ¼ + 1/6 + 1/8+ 1/12 = 19/12 < 2
Or, s <= 1, which is a contradiction.
Accordingly, p=1, giving:
(q-2)(r-2) =5, so that (q,r) = (3,7), giving:
(b,c) = (4,8)
or, (a,b,c) = (2,4,8)
Consequently, (a, b, c) = (2,4,8); (3, 5, 15) gives all possible solutions to the given problem.
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