Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Respected all Mathematicians! As we know that n-tuple is a set of n positive integers ($a_1$,...$a_n$) such that sum of the squares of this each member up to a_n-1 is square of $a_n$.

If we have (n-2) numbers out of n, we can find those and form a n-tuple by considering the following:

square of $a_1$ + square of $a_2$ + ... + square of a_n-2 = k and $a_n$ - a_(n-1) = S.

by considering the above cited information, I will take $a_1$ = 55, $a_2$ = 15, $a_3$ = 20, $a_4$= 10, $a_5$= 35, $a_6$= 45, $a_7$= 30 and $a_8$ = 25. I consider n = 10 and known n-2 = 8 values are given above. Now I will find the other two values $a_9$ and $a_10$.

Clearly, k = $5^2$ X 11 X 31 = 8525 and S = 1, $5^2$, 11, 31

Now possible values of S are $5^0$ $11^0$ $31^0$ = 1, $a_9$ = 4262 and the $a_10$ = 4263. Again, S = $5^0$ $11^1$ $31^0$ = 11 and $a_9$ = 382 and $a_10$ = 393 Again, S = $5^2$ $11^0$ $31^0$ = 25 and $a_9$ = 158 and $a_10$ = 183. Finally, S = $5^0$ $11^0$ $31^1$ = 31 and $a_9$ = 122 and $a_10$ = 153.

So, the primitive n-tuples ( n= 10) for know 8 values are

($55^2$ + $15^2$ + $20^2$ + $10^2$ + $35^2$ + $45^2$ + $30^2$ + $25^2$) + $4262^2$ = $4263^2$. Similarly we can check the other above cited values to conclude this method. I am looking for generalization of this method. Thanking you, BABA

share|improve this question
1  
I assume I am seeing it the tenth time! –  user21436 Feb 11 '12 at 7:50

1 Answer 1

I have difficulty understanding your question; its notation is somewhat unconventional, and I'm not sure what method you're asking us to generalize.

In general, if $n$ is an odd number, then $x + \left (\frac{x - 1}{2} \right)^2 = \left (\frac{x + 1}{2} \right)^2$. Therefore, if $x = a_1^2 + a_2^2 + \cdots + a_{n-1}^2$ is an odd number, then $(a_1, a_2, \ldots, a_{n-1}, \frac{x - 1}{2}, \frac{x + 1}{2})$ is a Pythagorean tuple. Even more generally, if $x$ can be written as $(a - b)(a + b)$, then $x + b^2 = a^2$.

Does this help?

share|improve this answer

Your Answer

 
discard

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.