Sums that are pythagorean and normal

Question

I noticed that

$3^2+4^2+15^2=9^2+13^2$

and also

$3+4+15=9+13$

Is there an easy way to find all pairs of disjoint sets of positive integers whose sum are the same and whose sum of squares are the same? How common are they?

Answer 1

It is easy to generate two such sets of arbitrary length. Let us choose $s_1,s_2,\ldots,s_{n-1}$ and $t_1,t_2,\ldots,t_{m-1}$ fairly freely and write $$ \begin{array}{cc} \sigma_1 = \sum_{i=1}^{n-1} s_i &~& \sigma_2 = \sum_{i=1}^{n-1} s_i^2 \\ \tau_1 = \sum_{i=1}^{m-1} t_i &~& \tau_2 = \sum_{i=1}^{m-1} t_i^2 \\ \end{array} $$ and then try to see if we can choose $s_n,t_m$ to augment our lists and satisfy the desired criterion, that is $$ \begin{align} \sigma_1-\tau_1 &= t_m-s_n \\ \sigma_2-\tau_2 &= t_m^2-s_n^2 \\ &=(t_m-s_n)(t_m+s_n) \\ &=(\sigma_1-\tau_1)(t_m+s_n) \end{align} $$ The easiest case is if $|\sigma_1-\tau_1|=1$ and $\sigma_2-\tau_2$ is odd, then we can choose $s_n,t_m=\frac{1}{2}(|\sigma_2-\tau_2|\pm 1)$.

For example for $n=11,m=8$, let $\{s_i\}_{i=1}^{10} = \{1,2,3,5,8,13,21,34,55,89\}$, $\{t_i\}_{i=1}^7=\{4,9,16,25,36,49,93\}$ where we have picked $t_7=93$ to achieve $\tau_1=232=\sigma_1+1$ because we will have $\tau_2>\sigma_2$.

Then $\tau_2-\sigma_2 = 509$, so let $s_{11}=255,t_8=254$ to get

$$\small 1+2+3+5+8+13+21+34+55+89+255=4+9+16+25+36+49+93+254=486 $$ $$ 1^2+2^2+3^2+5^2+8^2+13^2+21^2+34^2+55^2+89^2+255^2 \\ =4^2+9^2+16^2+25^2+36^2+49^2+93^2+254^2 = 77840 $$

In order to get every solution you'll have to consider any possible value of $d=|\sigma_1-\tau_1|$, and sets for which at least

• $d$ divides $|\sigma_2-\tau_2|$
• $(\sigma_2-\tau_2)/d \equiv d\pmod{2}$, since they sum to $2t_m$
• after adding $s_n,t_m$ the sets are still disjoint

and if you want to count without repetition then only include new solutions when $s_n$ and $t_m$ are the maximums of their respective sets.

Answer 2

I don't know that this is a complete parametrization, but if one takes any $y,z$ and then puts $x=(1+y)(1+z)$ and also $a=y+z+yz$ and $b=1+y+z$, one has $$x+y+z=a+b,\\ x^2+y^2+z^2=a^2+b^2.$$ Of course one can permute the formulas and get others; maybe someone can show this list is complete (I'll try for that). By the way I used maple12's "isolve" to get this parametric solution, and in general isolve is not guaranteed to get all solutions.