Magyar Information Contest Journal Articles

# Problem C. 1051. (November 2010)

C. 1051. A natural number n is chosen between two consecutive square numbers. The smaller square is obtained by subtracting k from n, and the larger one is obtained by adding l to n. Prove that the number n-kl is a perfect square.

(5 pont)

Deadline expired on December 10, 2010.

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

Megoldás. A két egymást követő négyzetszám legyen $\displaystyle x^2$ és $\displaystyle (x+1)^2$. Így a feladat szerint $\displaystyle n, k, l$-re $\displaystyle x^2+k=n$ és $\displaystyle n+l=(x+1)^2$. Kifejezve $\displaystyle k$-t és $\displaystyle l$-t $\displaystyle k=n-x^2$ és $\displaystyle l=(x+1)^2-n$ a vizsgálandó kifejezés $\displaystyle n-kl=n-(n-x^2)((x+1)^2-n)=n-(n(x+1)^2)-n^2-x^2(x+1)^2+nx^2)=n-n(2x^2+2x+1)+n^2+(x(x+1))^2=(x(x+1)-n)^2$, ami valóban négyzetszám.

### Statistics:

 211 students sent a solution. 5 points: 168 students. 4 points: 22 students. 3 points: 4 students. 2 points: 4 students. 1 point: 7 students. 0 point: 2 students. Unfair, not evaluated: 3 solutions. Unfair, not evaluated: 1 solution.

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