Magyar Information Contest Journal Articles

# Problem C. 1219. (March 2014)

C. 1219. Prove that 9m (where m is a positive integer) can always be expressed as the sum of three positive square numbers.

(5 pont)

Deadline expired on 10 April 2014.

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

Megoldás. Ha $\displaystyle m=1$, akkor $\displaystyle 9^1=2^2+2^2+1^2$.

Ha $\displaystyle m\geq2$, akkor ennek felhasználásával:

$\displaystyle (2\cdot3^{m-1})^2+(2\cdot3^{m-1})^2+(1\cdot3^{m-1})^2=4\cdot9^{m-1}+4\cdot9^{m-1}+1\cdot9^{m-1}=$

$\displaystyle =(4+4+1)\cdot9^{m-1}=9^m.$

### Statistics:

 136 students sent a solution. 5 points: 129 students. 4 points: 1 student. 3 points: 3 students. 2 points: 2 students. 1 point: 1 student.

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