Problem K. 318. (December 2011)
K. 318. Prove that if a, b, c, d are consecutive natural numbers, then d^{2} is a factor of the sum a+b^{2}+c^{3}.
(6 pont)
Deadline expired on 10 January 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.
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