perplexus.info :: Just Math : Some Squares Sum Year

Some Squares Sum Year (Posted on 2014-01-16)

Each of x, y and z is a positive integer satisfying:
x²+y²+z² = 2014

How many solutions are there? For which triplets is x+y+z a perfect square? How many distinct values can x+y+z have?

*** Disregard permutations. For example, (p,q,r) and (p,r,q) should be treated as the same solution.

Submitted by K Sengupta

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Solution: (Hide)

There are 9 distinct solutions of (x, y, z), and these are:
(x, y, z) = (3, 18, 41), (3,22,39), (5,15,42), (5, 30, 33), (9,13,42), (13,18,39), (14,27,33),(18,27,31), (21,22,33)
For (x,y,z) = (3,22,39), and: (9,13,42), we have:
x+y+z = 8²
The 6 distinct sums are 62, 64, 68, 70, 74, 76
Since 2014 is even but not divisible by 4, every triplet consists of two odds and one even. This is why x+y+z is always even.
For an explanation, refer to the respective solutions submitted by Jer and Charlie in the comments.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: I can program -- same answers in Basic	Charlie	2014-01-16 15:53:34
I can program	Jer	2014-01-16 15:38:29

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