Problem K/C. 888. (January 2026)

K/C. 888. The sum of the squares of three consecutive odd integers is a four-digit number with four equal digits. Find all such triples of positive integers.

(5 pont)

Deadline expired on February 10, 2026.

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

Megoldás. \(\displaystyle (2n-1)^2+(2n+1)^2+(2n+3)^2=12n^2+12n+11 = 1111\) vagy \(\displaystyle 2222\) vagy \(\displaystyle \ldots\) vagy \(\displaystyle 9999\). Vonjuk ki a \(\displaystyle 11\)-et, így \(\displaystyle 12n(n+1) = 1100\) vagy \(\displaystyle 2211\) vagy \(\displaystyle \ldots\) vagy \(\displaystyle 9988\). Az előző számok közül egyedül az \(\displaystyle 5544\) osztható \(\displaystyle 12\)-vel, tehát \(\displaystyle 12n(n+1) = 5544\), ahonnan \(\displaystyle n = 21\), a keresett számok pedig \(\displaystyle 41\), \(\displaystyle 43\) és \(\displaystyle 45\).

Statistics:

232 students sent a solution.
5 points:	184 students.
4 points:	28 students.
3 points:	10 students.
2 points:	3 students.
1 point:	3 students.
0 point:	1 student.
Unfair, not evaluated:	1 solutions.
Not shown because of missing birth date or parental permission:	1 solutions.

Problems in Mathematics of KöMaL, January 2026