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 $a+4=7$
 A: Hint.. $a+4\equiv a+1\quad \pmod 3$
A: HINT: One of the numbers $a,a+2$, and $a+4$ must be divisible by $3$. Why?
A: 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 $2,4,8,10,14,20,\cdot\cdot\cdot$.
A: Hint $\ $ They're odd so $\equiv 1,3,5\pmod 6$ so the one $\equiv 3$ must be $= 3,$ being prime.
A: $a\equiv 0 \mod 3\Rightarrow a=3$
$a\equiv 1\mod 3\Rightarrow a+2\equiv 0\mod 3\Rightarrow a+2=3\Rightarrow a=1$
$a\equiv 2\mod 3\Rightarrow a+4\equiv0\mod 3\Rightarrow a+4=3\Rightarrow a=-1$
So the only possibility is the first one.
A: $a$,$a+1$,$a+2$,$a+3$ and $a+4$ is a set of five consecutive numbers, $a \gt 3$.
any set of five consecutive numbers for $a \gt 3$, must consist of $two$ odd and $three$ even numbers, or $three$ odd and $two$ even numbers.
one just needs to consider the instance, when the set of five consecutive numbers consists of $three$ odd, and $two$ even numbers.
Among this set of five consecutive numbers, a maximum of $two$ of the numbers must be divisible by $3$ and they must neither be both odd, nor be both even.
we see that, when we have just one of the number which is divisible by $3$, this number will be odd and there will always be $two$ other even numbers, as we have a set which consists of $three$ odd and $two$ even numbers.
further, when, we have two of the numbers which are divisible by $3$, one of these numbers will be even and there will always be another even number, as we have a set which consists of $three$ odd and $two$ even numbers.
hence we can conclude that, for $a \gt 3$, no set of five consecutive numbers exists which consists of $three$ prime numbers.
A: $a$ is odd (why$?$)
one of $a,a+2,a+4$ is div. by $3$ and these three being prime $\implies$ one of them is $3$.
Since $3$ is the least odd prime and $a$ is the smallest among these three primes $\implies a=3$ is the only possibility and hence only one solution. 
A: You have observed that $a=3$ is a possible solution. Now assume $a>3$. What forms can $a$ take? Since $a$ is prime it should be of the form $3k+2$ or $3k+1$ where $k$ is a positive integer. If it was of the form $3k+2$ then $a+4=3k+2+4=3(k+2)$ will not be prime since $k+2 >1 $. If it was of the form $3k+1$ then $a+2=3k+1+2=3(k+1)$ will not be a prime since $k+1>1 $. Therefore, $k=3$ is the only answer. 
