(05-10-2015 05:11 PM)Gerald H Wrote: [ -> ]Yes, repeated elements & zeros permitted, also negative numbers - zeros only in the evens list!
Bravo on finding a solution - the first & only so far, but here's one more
{ 31109 355553 355553 26669 }
{ 384442 326664 2220 2220 }
Thanks! Just a little explanation on how I did it:
Let a, b, c, d, e, f, g and h be positive integer numbers. Then
2a, 2b, 2c and 2d are even numbers and
(2e + 1), (2f + 1), (2g + 1) and (2h + 1) are odd numbers
Thus, a² + b² + c² + d² = (2e + 1)² + (2f + 1)² + (2g + 1)² + (2h + 1)²
...
.
.... a² + b² + c² + d² = e² + f² + g² + h² + e + f + g + h + 1
So, in order to find pairs of list it suffices to choose four arbitrary integers, e, f, g and h, and solve the equation. For instance, e = 15, f = 10, g = 18 and h = 23. Then
a² + b² + c² + d² = 15² + 10² + 18² + 23² + 15 + 10 + 18 + 23 + 1
a² + b² + c² + d² = 1245
Lagrange's four-square theorem says that every natural number can be represented as the sum of four integer squares. This means there will always be integer solutions for the equation above, but finding them is a problem. In this case, one solution is a = 2, b = 4, c =21 and d = 28 (
WolframAlpha lists them all).
Thus, the list of even and odd numbers are, respectively:
{ 4 8 42 56 } and { 31 21 37 47}
4² + 8² + 42² + 56² = 31² + 21² + 37² + 47² = 4980
This doesn't appear to be a promising method, though.
Edited to fix a typo