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:

210 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 solutionss.

Problems in Mathematics of KöMaL, November 2010