Quiet Week-end Challenge (49G?) - Closed

[Gerson W. Barbosa]

(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