And one more similar one…

Take any number n_1>1. It’s divisible by at least one prime. Since it has no common factors with n_1+1, then the new number n_2=n_1(n_1+1) is divisible by at least two primes. Then n_3=n_2(n_2+1) is divisible by at least 3 primes by similar reasoning. In general, n_k is divisible by at least k distinct primes. Since k can grow arbitrarily large, there must be arbitrarily large primes.

This proof has a similar vibe to the first two I gave. In fact, if you unravel the topological proof I gave yesterday, you’ll see even that has its similarities. I promise, we’re going to move on to genuinely different things now.

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One Response to And one more similar one…

  1. Pingback: No proof today « Andy Soffer

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