a perfect square is never what?
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Assume, n≥4n4. By Bertrand's postulate there is a prime, let's call it pp such that n2<p<nn2pn . Suppose, p2p2 divides nn. Then, there should be another number mm such that p<m≤npmn such that pp divides mm. So, mp≥2mp2, then, m≥2p>nm2pn. This is a contradiction. So, pp divides n!n but p2p2 does not. So n!n is not a perfect square.
http://en.wikipedia.org/wiki/Bertrand_postulate
That leaves two more cases. We check directly, 2!=222 and 3!=636 are not perfect squares.
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http://en.wikipedia.org/wiki/Bertrand_postulate
That leaves two more cases. We check directly, 2!=222 and 3!=636 are not perfect squares.
share improve this answerk.
Example my dad main diet
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