 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Find the triplets (Posted on 2012-12-04) Find all triplets(x,y,z) of integers such that
1/x2 + 1/y2 + 1/z2 = 2/3

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Is this the solution? | Comment 1 of 2

WLOG, x, y and z can all be positive with x<=y<=z, as they can be permuted and negated in any combination without changing the value.

Since x is the smallest integer, 1/x^2 is the largest term. It can't be 1 as that would already exceed 2/3. But if x is 3 or larger, then 1/x^2 would be 1/9 or smaller, and since this is the largest term, the total couldn't be larger than 3/9 = 1/3.

So the smallest integer, which we have assigned to x, must be 2.

Then 1/y^2 + 1/z^2 must equal 2/3 - 1/4 = 5/12.

Again, since y <= z, y can't be 3 or more as 1/9 + 1/9 would fail to be as high as 5/12. So y could only be 2.

But then the third term, 1/z^2, would have to be 2/3 - 1/4 - 1/4 = 1/6, which is not the reciprocal of a perfect square.

Therefore, there can be no set of integers that satisfies the equation.

 Posted by Charlie on 2012-12-04 14:36:19 Please log in:

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