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Let $a<b<c$ be primes such that $c-a$, $c-b$, and $b-a$ are also prime. It is rather simple to show that $(2,5,7)$ is the only triple that satisfies these conditions:

Proof Sketch:

  1. The case $a>2$ reduces to a system of equations with no solution after realizing that each difference must be equal to $2$.

  2. The case $a=2$ reduces to the existence of a prime $p$ such that $p+2$ and $p+4$ are also prime. The only such tuple is $p=3\rightarrow (3,5,7)$. A modular argument takes care of uniqueness.

  3. The tuple $(a,b,c)=(2,5,7)$ follows from $p=3$.

I was wondering if there is a more elegant approach utilizing many number theoretic tools (i.e. elliptic curves, algebraic number theory, ect.)? I realize that this is entirely unnecessary as we can appeal to the most atomic of theory to solve this problem; however, I don't spend much time with number theory and was looking for some application of "modern techniques".

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  • $\begingroup$ Probably not. Sorry. $\endgroup$ – Qiaochu Yuan Sep 27 '15 at 2:30
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    $\begingroup$ Care to elaborate? As I stated, I have little "feel" for such techniques. $\endgroup$ – Joseph Zambrano Sep 27 '15 at 2:32
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    $\begingroup$ @QiaochuYuan Really. Op writes an entire post, well made too, and you respond with an opinion worth ten cents? I mean, what is the point in doing that? You don't even say it can't be done you just vacillate about the possibility of being wrong... $\endgroup$ – Zach466920 Sep 27 '15 at 4:34
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    $\begingroup$ @Zach466920: look, it's a comment specifically because it's not an answer. It's very hard to rule out the possibility of a technique being applied to a problem, but it's unlikely in this case. Problems of this general type belong to an area sometimes called "additive number theory" and the techniques there just look very different from the techniques used in algebraic number theory, etc. In any case, there's a perfectly nice elementary solution that makes the problem very clear; I don't see a need to go hunting for more sophisticated techniques on this particular problem relative to others. $\endgroup$ – Qiaochu Yuan Sep 27 '15 at 5:01
  • $\begingroup$ @QiaochuYaun Thank you for the clarification. I will certainly look into some additive number theoretic techniques. To clarify, this isn't an attempt to hunt for overly complicated techniques; rather, I am interested in juxtaposing the elementary proof with something "higher level" to see what such techniques have to offer (perhaps some deeper context). Like I said, I have spent little time in this domain so I thought this might be a good example to do just that. $\endgroup$ – Joseph Zambrano Sep 27 '15 at 13:24
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So I actually not giving the mathematical proof but I am describing the reason of not having more triplets.

As a$<$b$<$c are primes such that c-a, c-b and b-a are also primes.

So we can't take all primes as odd one because there does't exist such odd primes a,b&c which will satisfy our condition because the values of c-a, c-b and b-a are even numbers and one of these values must greater than 2, which will not be a prime.

Now it is clear that out of these 3 primes (a,b&c) one should be 2 and it must be 'a' than only we can get 'b' and 'c' greater than 'a'.

Suppose if we take 'b' & 'c' as a pair of twin primes and 'a' as 2 than (c-a) is must be prime, (c-b) must be prime but it is not necessary that (b-a) is also prime because as (a=2) so b-2 is not prime as 'b' & 'c' are already twin primes so 'b-2' can't be a twin prime with 'b' except (b=5) therefore 'b-a' or 'b-2' is not a prime number (exception, b=5)

Therefore there is only 1 triplet of your conditions.

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    $\begingroup$ "it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3. $\endgroup$ – eyeballfrog Dec 18 '18 at 17:33
  • $\begingroup$ @eyeballfrog I edited my answer, is it now nearby correct answer or not? $\endgroup$ – Dynamo Dec 18 '18 at 17:42
  • $\begingroup$ That works, but it's dancing around the point a little. The point is that one of $b -2,\,b\,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3. $\endgroup$ – eyeballfrog Dec 18 '18 at 17:46
  • $\begingroup$ @eyeballfrog Yes, you are right... $\endgroup$ – Dynamo Dec 18 '18 at 17:47

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