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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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 Diophantine Thoughts | Comment 4 of 5 |
(In reply to Solving tips by Brian Smith)

To avoid confusion with the x in a+b=x^2 and a = k*(x^2-y^2) ......... :

we will substitute (a,b,y) = (k*(p^2-q^2), k*(2*p*q), k*p^2+q^2), where each of p, q and r are positive integers.......(*) in the first equation (a+b = x^2), and arrive at the following relationship after some manipulation.

(p, q, x) = (k(m^2-n^2), 2*k*n(m+n),k(m^2-n^2 -2*m*n)), where a, b and c are as given in (*) ...... Form (I)

Alternatively, taking g = 2k, and (m, n) = (s+t, s-t), we get the
form:

(p, q, x) = (g(s^2+ t^2), 2*g*s(s-t),g(s^2-t^2 -2*s*t))
...... Form (II). Note that q  may be positive or negative accordingly, as s> t or s< t assuming that each of s and t are positive integers.

Since, it is easier to work with the second form, WLOG we can assume that g, s and t are positive integers and k is a positive integer or positive half integer.

Thus, form (II) gives the primitive pairs (a, b) in  conformity with the given equations (i) and (ii). We now have to check the given equation (iii), to ascertain which of these pairs (a, b) satisfies a^3+b^3=z^2, where z is a positive integer.

*** I am not fully au fait with the various computer programs, and have been unsuccessful in my attempts by hand to derive any valid pair (a, b) in conformity with all the three equations apart from the example given in the problem text.

Edited on April 24, 2008, 6:58 am
 Posted by K Sengupta on 2008-04-24 06:08:08

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