Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem C. 1401. (February 2017)

C. 1401. The positive numbers $\displaystyle x$, $\displaystyle y$ satisfy the equation $\displaystyle x^3+y^3=x-y$. Prove that $\displaystyle x^2+y^2<1$.

(5 pont)

Deadline expired on March 10, 2017.

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

Megoldás. Mivel $\displaystyle x$ és $\displaystyle y$ pozitív számok, így

$\displaystyle x^3+y^3=x-y=\frac{(x-y)(x^2+xy+y^2 )}{x^2+xy+y^2}=\frac{x^3-y^3}{x^2+xy+y^2}.$

Átrendezve:

$\displaystyle x^2+xy+y^2=\frac{x^3-y^3}{x^3+y^3}.$

Ebből következik, hogy

$\displaystyle x^2+y^2<x^2+xy+y^2=\frac{x^3-y^3}{x^3+y^3}<1,$

mert a tört számlálója kisebb, mint a nevezője.

Tehát $\displaystyle x^2+y^2<1$.

### Statistics:

 119 students sent a solution. 5 points: 96 students. 4 points: 6 students. 3 points: 4 students. 2 points: 5 students. 0 point: 6 students. Unfair, not evaluated: 2 solutions.

Problems in Mathematics of KöMaL, February 2017