Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem B. 3893. (March 2006)

B. 3893. Solve the equation

(x2+y2)3=(x3-y3)2

on the set of real numbers.

(3 pont)

Deadline expired on April 18, 2006.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás: A zárójeleket kibontva, rendezés után a 3x4y2+3x2y4+2x3y3=0 egyenletre jutunk, vagyis

x^2y^2\bigl(2x^2+2y^2+(x+y)^2\bigr)=0.

Ez pedig pontosan akkor teljesül, ha x és y közül legalább az egyik 0.


Statistics:

204 students sent a solution.
3 points:154 students.
2 points:19 students.
1 point:28 students.
0 point:2 students.
Unfair, not evaluated:1 solutions.

Problems in Mathematics of KöMaL, March 2006