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What is the next triplet in the below series of triplets:

(2,3,4), (47,48,49), (674,675,676), (9407,9408,9409), ___?

Few points about the above series:

(i) the terms of the triplets are consecutive

(ii) there is at least one square among the three terms

(iii) the sum of the terms of each triplet is also a square

I tried to find an underlying series/formula for finding such triplets, but in vain.

So, can you please help me out!

inavda
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Karri Chandrasekhar
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    Thanks for down voting, without revealing the reason. – Karri Chandrasekhar Mar 26 '19 at 16:38
  • It seems reasonable to assume the next triple will consist of consecutive numbers. Which ones is essentially arbitrary; I can concoct a formula for justifying any, say, middle term. – egreg Mar 27 '19 at 11:00

1 Answers1

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We are basically searching all the triples $(x-2,x-1,x)$ such that:

$$(x-1)+(x-2)+x=3x-3=\beta^2 \text{ and } x=\alpha^2$$

1 Condition

$$3x-3=\beta^2 \Rightarrow 3(x-1)=\beta^2$$

If $3|\beta^2$ then $9|\beta^2$ so $x-1=3k^2$. So the triples we are searching are of this form $(3k^2-1,3k^2,3k^2+1)$

2 Condition

We must now impose that:

$$x=\alpha^2 \Rightarrow \alpha^2-3k^2=1 $$

Which is a standard Pell's equation. Solving it with Wolfram Alpha(I can do this alone but I haven't time ) we find that the first solution are:

$$k=1$$$$k=4$$$$k=15$$$$k=56$$$$k=209$$

The first 4 work and we find again the given triples. To find the next one we must consider $k=209$ that gives in output:

$$(131042,131043,131044)$$

:)

Moreover searching on OEIS it seems that the central number of each triples has a particular propriety connected with standard deviation: http://oeis.org/A007654

Kandinskij
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