Problem K. 318. (December 2011)

K. 318. Prove that if a, b, c, d are consecutive natural numbers, then d² is a factor of the sum a+b²+c³.

(6 pont)

Deadline expired on January 10, 2012.

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

Megoldás. Az első három szám legyen \(\displaystyle d–3\), \(\displaystyle d–2\), \(\displaystyle d–1\). Ekkor az összeg: \(\displaystyle d–3 + (d–2)^2 + (d–1)^3 = d – 3 + d^2 – 4d + 4 + d^3 – 3d^2 + 3d – 1 = d^3 – 2d^2 = d^2(d–2)\), azaz osztható \(\displaystyle d\)–vel.

Statistics:

211 students sent a solution.
6 points:	139 students.
5 points:	17 students.
4 points:	14 students.
3 points:	9 students.
2 points:	6 students.
1 point:	5 students.
0 point:	5 students.
Unfair, not evaluated:	16 solutionss.

Problems in Mathematics of KöMaL, December 2011