<s>[INST] If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$? If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$?
when, $a=3$
we have, $a+2=5$ and $a+4=7$
 [/INST] First of all $a$ must be odd.
If prime $a>3,$ it must be either $6b+1$ or $6b-1$, where $b$ is a natural number  $≥1$.
If $a=6b+1, a+2=3(2b+1)$ is composite as $2b+1≥3$
If $a=6b-1, a+4=3(2b+1)$ is composite as $2b+1≥3$
In fact, $3\mid a(a+k)(a+2k)$ where $k$ is positive integer with $(3,k)=1$
As $a(a+k)(a+2k)=a^3+3a^2k+2ak^2≡a^3+2ak^2\pmod 3≡a^3+2a$ as $k^2≡1\pmod 3$
So,$a(a+2k)(a+4k)≡a^3+2a\pmod 3≡a(a-1)(a+1)+3a\pmod 3$
So if $a>3$ and $(3,k)=1$, one of $a, (a+k)$ or $(a+2k)$ is divisible by $3$, hence is composite.
Observe that exactly one of them is divisible by $3$.
So, if $a≠3$, all of $a,a+k,a+2k$ can not be prime.
Again, $k$ must be even to keep $a+k,a+2k$ odd.
So, $k$ must be of the form $6m±2$  as $(3,k)=1$.
By observation, some of the values of $k$ for which all of $3,3+k,3+2k$ are prime, are $2,4,8,10,14,20,\cdot\cdot\cdot$.
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