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Source: Peru EGMO TST 2018

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#1 h Feb 13, 2019, 12:49 AM • 2 Y

Y by Adventure10, Mango247

The number $n$ is "good", if there is three divisors of $n$ ( $d_1, d_2, d_3$ ), such that $d_1^2+d_2^2+d_3^2=n$
a) Prove that all good number is divisible by $3$
b) Determine if there are infinite good numbers.

This post has been edited 1 time. Last edited by mathisreal, Feb 13, 2019, 12:50 AM

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#2 h Feb 13, 2019, 3:19 AM • 2 Y

Y by mathisreal, Adventure10

(a)
Assume that $n=d^{2}_{1} +d^{2}_{2} +d^{2}_{3}$ and $n$ is not divisible by $3$ .
Then $d_{i}$ is not divisible by $3$ .
So $d_{i} \equiv \pm 1\pmod 3$ for $i=1,2,3$ .
Note that $n-3=( d_{1} +1)( d_{1} -1) +( d_{2} +1)( d_{2} -1) +( d_{3} +1)( d_{3} -1)$ .
Since $\text{RHS}$ is divisible by $3$ , we have $\text{LHS}$ is also divisible by $3$ .
It follows that $n$ is divisibly by $3$ , which is a contradiction.

(b)
$n=30k^{2} =k^{2} +( 2k)^{2} +( 5k)^{2}$

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