Let p 1 = 2 , p 2 = 3 , p 3 = 5 , p 4 = 7 , . . . be the primes. Compute ( 6 5 5 3 7 − p 1 ) ( 6 5 5 3 7 − p 2 ) . . . ( 6 5 5 3 7 − p 1 0 0 0 0 ) m o d 6 5 5 3 7
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There's a simpler way to prove that the 10000th prime number is larger than 65537.
Hint : Prime Number Theorem.
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Note that 6 5 5 3 7 = 2 1 6 + 1 , which is a well-known Fermat prime. A quick search will reveal that there are 9 5 9 2 primes less than 1 0 0 0 0 0 , so one of the primes used will be 6 5 5 3 7 , which makes the whole product 0 .