# The prime 13 (TOMATO 37)

Supposed p is a prime Number such that (p-1)/4 and (p+1)/2 are also primes. Show that p=13.

Discussion:

p is not 2 or 3 (otherwise (p-1)/4 would not be an integer).Hence p must be an odd prime. Also p-1 is divisible by 4

p = 4t + 1 (say)

(p-1)/4 = t

(p+1)/2 = 2t + 1

Hence t, 2t+1, 4t+1 are all primes.

If t = 3 then these numbers are 3, 7 and 13.

If t is not 3 then t must produces 1 or 2 remainder when divided by 3 (t is a prime, hence cannot be divisible by 3).

$$\displaystyle {t \equiv 1 \text{mod} 3 \implies 2t +1 \equiv 3 \equiv 0 \text {mod} 3}$$

But 2t +1 is a prime. So it is impossible that 3 divides 2t +1. Hence t cannot be 1 mod 3.

Similarly $$\displaystyle {t \equiv 2 \text{mod} 3 \implies 4t +1 \equiv 9 \equiv 0 \text {mod} 3}$$

But 4t +1 is a prime. So it is impossible that 3 divides 4t +1. Hence t cannot be 1 mod 3.

Therefore we can have no other t such that the given condition is satisfied. Hence t must be 3 and the prime must be 13.