Description: Lemma for 4sq . The proof is by strong induction - we show that if all the integers less than k are in S , then k is as well. In this part of the proof we do the induction argument and dispense with all the cases except the odd prime case, which is sent to 4sqlem18 . If k is 0 , 1 , 2 , we show k e. S directly; otherwise if k is composite, k is the product of two numbers less than it (and hence in S by assumption), so by mul4sq k e. S . (Contributed by Mario Carneiro, 14-Jul-2014) (Revised by Mario Carneiro, 20-Jun-2015)
Ref | Expression | ||
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Hypothesis | 4sq.1 | |
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Assertion | 4sqlem19 | |