Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a+4=7$

share|cite|improve this question
Funny question! (+1) :-) – user 1618033 Aug 31 '12 at 18:16

HINT: One of the numbers $a,a+2$, and $a+4$ must be divisible by $3$. Why?

share|cite|improve this answer
Good answer (+1) – user 1618033 Aug 31 '12 at 18:15
On the border of begin outrageous, good comments!! – Jayesh Badwaik Sep 1 '12 at 3:57
It's a shame that the twin primes hypothesis can't be proved or dismissed so simply. – Cameron Buie Sep 8 '12 at 4:08

$a\equiv 0 \mod 3\Rightarrow a=3$

$a\equiv 1\mod 3\Rightarrow a+2\equiv 0\mod 3\Rightarrow a+2=3\Rightarrow a=1$

$a\equiv 2\mod 3\Rightarrow a+4\equiv0\mod 3\Rightarrow a+4=3\Rightarrow a=-1$

So the only possibility is the first one.

share|cite|improve this answer

Hint.. $a+4\equiv a+1\quad \pmod 3$

share|cite|improve this answer

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 $2,4,8,10,14,20,\cdot\cdot\cdot$.

share|cite|improve this answer

Hint $\ $ They're odd so $\equiv 1,3,5\pmod 6$ so the one $\equiv 3$ must be $= 3,$ being prime.

share|cite|improve this answer

$a$ 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.

share|cite|improve this answer

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.

share|cite|improve this answer

$a$,$a+1$,$a+2$,$a+3$ 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.

share|cite|improve this answer
CAn you explain the bit where you say "two of the numbers must be divisible by 3"? I don't follow that step of the logic... – Chris Aug 31 '12 at 15:47
What about 7,8,9,10,11. This has three odd numbers and only one divisible by 3... – Chris Aug 31 '12 at 22:02
a set of five consecutive numbers greater than 3, with three odd numbers (the first, the middle and the last numbers of the set are odd) will be of the form, n+4,n+5,n+6,n+7 and n+8, where n is an odd natural number. On the number line every 3rd number after 0 is divisible by 3, e.g. 3,6,9,12,15,⋯,2n+1, i.e. in the above set of five consecutive numbers, there must be one odd number which will be divisible by 3. – Rajesh K Singh Sep 1 '12 at 2:58
please, see the latest edition. – Rajesh K Singh Sep 1 '12 at 5:35
Ah, I think I understand your logic now. – Chris Sep 1 '12 at 10:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.