Pythagoras Extended

Pick whatever values of $a_1, a_2, \ldots, a_{n-1}$ you like so that $a_1^2 + \cdots + a_{n-1}^2 = m$ is odd and greater than $1.$ (For instance, you can take $a_1 = 3$ and $a_k = 2$ for $2 \le k \le n-1.$ ) Then the equation $m + a_n^2 = b^2$ always has solutions in positive integers, in particular $a_n = \frac{m-1}2, b = \frac{m+1}2.$

A well-known Pythagorean generator for $a^2 + b^2 = c^2$ is to pick positive integer values for $m$ and $n$ and use $a = m^2 - n^2$ , $b = 2mn$ , and $c = m^2 + n^2$ . This can concept be used over and over again to make longer equations with integer values.

One way is by using $m = k_1$ and $n = k_1 - 1$ for $a_1 = m^2 - n^2 = 2k_1 - 1$ , $b_1 = 2mn = 2k_1(k_1 - 1)$ , and $c_1 = m^2 + n^2 = 2k_1^2 - 2k_1 + 1$ , so that $a_1^2 + b_1^2 = c_1^2$ . Then we can set $c_1 = a_2$ , so that $a_2^2 + b_2^2 = c_2^2$ is also $a_1^2 + b_1^2 + b_2^2 = c_2^2$ . Then we set $c_2 = a_3$ , and so on, to keep extending the equation. Since $c_n = 2k_n^2 - 2k_n + 1$ , $a_{n + 1} = 2k_{n + 1} - 1$ , and $c_n = a_{n + 1}$ , this solves to:

$k_{n + 1} = k_n^2 - k_n + 1$

and:

$(2k_1 - 1)^2 + (2k_1(k_1 - 1))^2 + (2k_2(k_2 - 1))^2 + (2k_3(k_3 - 1))^2 + ... + (2k_{n - 1}(k_{n - 1} - 1))^2 = (2k_{n - 1}^2 - 2k_{n - 1} + 1)^2$

which satisfies the conditions for the given equation for every positive integer $n$ and any positive integer $k_1 \geq 2$ .

For example, using $k_1 = 2$ for $n = 5$ , the $k$ -values calculate to $2$ , $3$ , $7$ , and $43$ , which gives:

$3^2 + 4^2 + 12^2 + 84^2 + 3612^2 = 3613^2$

Take $a_1=a_2=...=a_n$ and $n=c^2$ , where $c$ is a positive integer. Then $a_1^2+a_2^2+...+a_n^2=na_1^2=c^2a_1^2=b^2$ , where $b=ca_1$ , a positive integer. (It is not told whether the $a_i$ 's are different or not. :))

Not a solution

Interesting article sparsely related to this. Very interesting