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 Sum of Powers (Posted on 2008-04-21)
The numbers 184 and 345 have a special property. Their sum, the sum of their squares, and the sum of their cubes are all perfect squares:
184 + 345 = 23^2
184^2 + 345^2 = 391^2
184^3 + 345^3 = 6877^2

Find another primitive pair of non-zero integers with the same property. Note, a primitive solution is a solution which is not a multiple of any smaller solution.
If you have extended precision math software, try to find a third or fourth primitive solution.

Tip: one of the numbers may be negative.

 No Solution Yet Submitted by Brian Smith Rating: 4.0000 (2 votes)

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 thoughts Comment 5 of 5 |
We can write a^3  +b^3 as (a+b)(a^2-ab+b^2). So if a+b is a perfect square and a^3+b^3 is a perfect square then so is (a^2-ab+b^2). So too is a^2+b^2

Suppose x and y are integers such that x^2 + y^2 = c^2 and x^2-xy+y^2 = d^2, but x+y is not a perfect square. Then create X = x(x+y) and Y = y(x+y). X+Y = (x+y)(x+y) which is a perfect square. X^2 + Y^2 = (x+y)^2(x^2+y^2) = (x+y)^2*c^2 which is also a perfect square. And X^3+Y^3 = (x+y)^3(x^3+y^3) = (x+y)^4(d^2) which is also a perfect square. (X,Y) is therefore a solution to the original problem.

So, if we can find x and y such that x^2+y^2 AND x^2-xy+y^2 are squares, we can construct a solution to the problem above even if x+y is not itself a square. Note that if x and y share a common factor z, then x/z and y/z are also solutions since both expressions above are then multiples of z^2 and the leftover must still be a perfect square. In this case, the transformed X and Y share a common factor of z^2 and the solution is non-primative. It appears primative solutions arise when x and y are relatively prime.

Actually finding another (x,y) from which to generate another solution is something I haven't yet managed to accomplish.

For the sample, x and y are 8 and 15. x^2+y^2 = 17^2 and x^2-xy+y^2 = 13^2. 184 = 8*(8+15) and 345 = 15*(8+15) and so on.

I don't have a solution, but at least this method allows the first equation to be ignored entirely in favor of solving two sets of equations in two unknowns.

 Posted by Paul on 2008-04-24 07:38:36

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