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Guest Message by DevFuse

# ABK

Member Since 14 Sep 2012
Offline Last Active Sep 14 2012 03:16 PM

### In Topic: Math - Prime number

14 September 2012 - 03:02 PM

My first approach was to consider modulo 6.

Assume a > 6, Since a is prime we have a=1, -1 (mod 6)

Thus, a2=1 (mod 6).

We know, 26 = 2 (mod 6)

Therefore a2+26=3 (mod 6)

So any prime greater than 6 would be resulting in a number which is divisible by 3 which NOT prime as required.

So we have to check 2, 3 and 5 which are the only primes less than 6. It is easy to see that each would yield a non-prime in a2+26. So we're done!