## Sum to Squares

Find three [unique positive] numbers, such that their sum is square, and the sum of any pair is a square.
• List all solutions with total sum less than 1000.
• Does a solution exist, such that the original three numbers are also square? For example, without the restriction that the numbers be positive, the set (0, 9, 16) would be a possible solution.

Source: Diophantos, as quoted in The Penguin Book of Curious and Interesting Puzzles, David Wells, 1992, #28.

Solutions were received from Philippe Fondanaiche, Denis Borris, Carlos Rivera, and Bill Chapp.

For a,b,c unique positive integers, and a+b+c < 1000

``` 41,  80, 320
57, 112, 672
88, 168, 273
```
The last question above does not have a simple solution. It is actually just a rewording of an as-yet-unsolved problem (I didn't realize this was the same quesiton when I wrote it.) Thanks to Philippe Fondanaiche for the following discussion:

W.Sierpinski in Elementary theory of numbers (editor:A.Schinzel North-Holland Mathematical Library):
"We don't know whether there exist three natural numbers a,b,c such the each of the numbers a^2 + b^2, a^2 + c^2, b^2 + c^2 and a^2 + b^2 +c^2 is the square of a natural number. In other words,we do not know whether there exists a rectangular parallelepiped whose sides,face diagonals and inner diagonal are all natutal numbers. At the opposite, there exist four natural numbers x,y,z,t such that the sum of the squares of any three of them is a square.
Euler's solution x = 168 y = 280 z = 105 t = 60
Tebay's solution x = 1995 y = 6384 z = 1520 t = 840 "

The three solutions above can actually be solved for analytically, but it's not obvious they are the only solutions for a total sum less than 1000. This is Diophantos' solution:
```Let a+b+c = (x+1)^2
Let   a+b = x^2      -->  c = 2x+1
Let   a+c = (x-1)^2  -->  b = 4x,  a = x^2 - 4x
Then we need merely find c+b = 6x+1 to be a square and we find our answer.
For x=20, 6x+1 = 121, and c=41, b=80, a=320.
For x=28, 6x+1 = 169, and c=57, b=112, a=672.
The third answer above is found by letting a+b+c=(x+2)^2 and a+c=(x-2)^2.
```

Here are Bill Chapp's simulation's answers.
```For a,b,c unique positive integers, and a+b+c < 1000

41,  80, 320
57, 112, 672
88, 168, 273

The full list of solutions for a,b,c postive integers < 1000:

17,  32,  32 Not unique
41,  80, 320
57, 112, 672
65, 464, 560 >1000
68, 128, 128 Not unique
72,  72, 217 Not unique
88, 168, 273
128, 128, 833 >1000, Not unique
136, 264, 825 >1000
144, 585, 640 >1000
153, 288, 288 Not unique
200, 200, 329 Not unique
272, 512, 512 >1000, Not unique
280, 345, 744 >1000
288, 288, 868 >1000, Not unique
385, 456, 840 >1000
425, 800, 800 >1000, Not unique
720, 801, 880 >1000
```

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