Problem K. 663. (September 2020)

K. 663. The sum of the squares of three consecutive integers equals the sum of the squares of the following two integers. What may be these five consecutive numbers?

(6 pont)

Deadline expired on October 12, 2020.

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

Megoldás.

\(\displaystyle (x – 2)^2 + (x – 1)^2 + x^2 = (x + 1)^2 + (x + 2)^2,\)

ahol \(\displaystyle x\) egész szám. A zárójeleket felbontva, majd az egyenletet rendezve:

\(\displaystyle 3x^2 – 6x + 5 = 2x^2 + 6x + 5,\)

\(\displaystyle x^2 – 12x = 0,\)

\(\displaystyle x(x – 12) = 0.\)

Azaz \(\displaystyle x = 0\) vagy \(\displaystyle x = 12\). A keresett számok: \(\displaystyle –2\), \(\displaystyle –1\), \(\displaystyle 0\), \(\displaystyle 1\), \(\displaystyle 2\) vagy \(\displaystyle 10\), \(\displaystyle 11\), \(\displaystyle 12\), \(\displaystyle 13\), \(\displaystyle 14\).

Statistics:

199 students sent a solution.
6 points:	96 students.
5 points:	22 students.
4 points:	20 students.
3 points:	18 students.
2 points:	4 students.
1 point:	26 students.
0 point:	8 students.
Not shown because of missing birth date or parental permission:	5 solutions.

Problems in Mathematics of KöMaL, September 2020