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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 $a+4=7$

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marked as duplicate by kingW3, Jack D'Aurizio elementary-number-theory Oct 15 '17 at 16:28

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    $\begingroup$ Funny question! (+1) :-) $\endgroup$ – user 1357113 Aug 31 '12 at 18:16
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HINT: One of the numbers $a,a+2$, and $a+4$ must be divisible by $3$. Why?

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    $\begingroup$ Good answer (+1) $\endgroup$ – user 1357113 Aug 31 '12 at 18:15
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    $\begingroup$ On the border of begin outrageous, good comments!! $\endgroup$ – user14082 Sep 1 '12 at 3:57
  • $\begingroup$ It's a shame that the twin primes hypothesis can't be proved or dismissed so simply. $\endgroup$ – Cameron Buie Sep 8 '12 at 4:08
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$a\equiv 0 \mod 3\Rightarrow a=3$

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

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

So the only possibility is the first one.

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

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

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$a$ 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.

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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.

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

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  • $\begingroup$ 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... $\endgroup$ – Chris Aug 31 '12 at 15:47
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    $\begingroup$ What about 7,8,9,10,11. This has three odd numbers and only one divisible by 3... $\endgroup$ – Chris Aug 31 '12 at 22:02
  • $\begingroup$ 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. $\endgroup$ – HOLYBIBLETHE Sep 1 '12 at 2:58
  • $\begingroup$ please, see the latest edition. $\endgroup$ – HOLYBIBLETHE Sep 1 '12 at 5:35
  • $\begingroup$ Ah, I think I understand your logic now. $\endgroup$ – Chris Sep 1 '12 at 10:10

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