For any integer n > 1, n and n+1 are coprime - mutually prime, having no common prime factors. So start with any n > 1 and write down one of its prime factors, say p. The prime factors of its successor, n + 1, are different from p. So there is at least some other prime, say q.

Now consider the successor of the product n(n + 1). The prime factors of the latter are different from those of n and n + 1, p and q, in particular. Let r be one of those.

Appply the same argument to the successor of n(n + 1)[n(n + 1) + 1] to obtain yet another prime, say s. Obviously the process can be extended indefinitely.

(I am grateful to Jens Bossaert for pointing out the source of the proof. The proof is due to Filip Saidak, a mathematician at the University of North Carolina.)

Reference

1. F. Saidak, A New Proof of Euclid's Theorem, The American Mathematical Monthly, Vol. 113, No. 10 (Dec., 2006), pp. 937-938

• Infinitude of Primes
  1. Infinitude of Primes - A Topological Proof
  2. Infinitude of Primes - A Topological Proof without Topology
  3. Infinitude of Primes Via *-Sets
  4. Infinitude of Primes Via Coprime Pairs
  5. Infinitude of Primes Via Fermat Numbers
  6. Infinitude of Primes Via Harmonic Series
  7. Infinitude of Primes Via Lower Bounds
  8. New Proof of Euclid's Theorem

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