discrete- mathematical proof to support the statement

Question

lets assume i have three children , tim,jack and juliet. currently tim is 14 years old , jack is 10 years and juliet is 6 years old. next year => tim:15, jack=11, juliet=7 and so on....

Is it possible that all of the three children will have an age that are a prime number, in the same year?

i know by checking the prime numbers list that it is not possible , but is there a mathematical proof(expression,equation,formula ) to have as an argument ? if there is not mathematical proof of that, what would your argument be? how would you convince me that it is not possible ?

Answer 1

If $6+n$, $10+n$ and $14+n$ are all primes, then $n$ must be odd. So we can let $n=2k+1$.

Hence we need to prove $2k+7$, $2k+11$ and $2k+15$ can not be primes simultaneously.

1. If we let $k=3p$, then $2k+15=6p+15=3(2p+5)$ is not prime.
2. If we let $k=3p+1$, then $2k+7=6p+9=3(2p+3)$ is not prime.
3. If we let $k=3p+2$, then $2k+11=6p+15=3(2p+5)$ is not prime.

Thus, the three number cannot be all primes.

Answer 2

It is impossible.

$14+2k,10+2k,6+2k\gt 2$ are even. So, we want three numbers $14+2k-1,10+2k-1,6+2k-1$ to be primes.

• For $k\equiv 0\pmod 3$, $10+2k-1\equiv 9\equiv 0\pmod 3$.

• For $k\equiv 1\pmod 3$, $14+2k-1\equiv 15\equiv 0\pmod 3$.

• For $k\equiv 2\pmod 3$, $6+2k-1\equiv 9\equiv 0\pmod 3$.

So, one of the three numbers is a multiple of $3$ and is larger than $3$, so is not a prime.