# n -tuples of pythagorean

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

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I assume I am seeing it the tenth time! –  user21436 Feb 11 '12 at 7:50

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 $x + b^2 = a^2$.