IMO Team Selection Test 1

Answer: $n=1$ and $n=6$.

We first note that $n=1$ satisfies.

We now prove that $\beta (n)=\frac{n}{2}$ for $n\ge 2$. Indeed, $\gcd (k,n)=\gcd (k-n,n)=\gcd (n-k,n)$, so $\gcd (k,n)=1$ if and only if $\gcd (n-k,n)=1$. This yields a partition of the positive integers $1\le k\le n-1$ with $\gcd (k,n)=1$ into pairs $(k,n-k)$, which have mean $\frac{n}{2}$. For $n\ge 2$, $\gcd (n,n)>1$, so we see that the average of all the numbers $1\le k\le n$ with $\gcd (k,n)=1$ is also equal to $\frac{n}{2}$. The number $\frac{n}{2}$ might have been double counted, but that does not matter since the average is $\frac{n}{2}$.

Prime numbers do not satisfy, because then $\alpha (n)=\frac{n+1}{2}$ and $\beta (n)=\frac{n}{2}$.

Now suppose $d$ is a positive divisor of $n$ with $d$ unequal to $1$ and $n$. Then we have $2\le d\le \frac{n}{2}$, so $(\frac{n}{2}-d)(d-2)\ge 0$, so $\frac{n}{2}d+2d\ge n+d^2$, so $\frac{n}{2}+2\ge \frac{n}{d}+d$.

Now suppose $n>6$ is compound. We can divide the positive divisors of $n$ into pairs $(d,\frac{n}{d})$. The pair $(1,n)$ has average $\frac{n+1}{2}$, all other pairs have average $\frac{\frac{n}{d}+d}{2}\le \frac{\frac{n}{2}+2}{2}<\frac{n-1}{2}$, or it is just the number $\sqrt{n}\le \frac{n}{3}<\frac{n-2}{2}$ (in this case, $n\ge 9$ holds). If $n$ is not the square of a prime number, we have a pair with average $\frac{n+1}{2}$, a pair with average smaller than $\frac{n-1}{2}$, and even possibly more pairs (or a single number) all with average smaller than $\frac{n-1}{2}$, making the total average smaller than $\frac{n}{2}$. If $n$ is the square of a prime number, we find $\alpha (n)=\frac{1+\sqrt{n}+n}{3}<\frac{1+\frac{n-2}{2}+n}{3}=\frac{3n}{6}=\frac{n}{2}$. So $n>6$ does not satisfy.

Also $n=4$ does not satisfy, because $\alpha (4)=\frac{1+2+4}{3}=\frac{7}{3}$ while $\beta (4)=\frac{4}{2}=2$.

Finally, we check that $n=1$ and $n=6$ indeed satisfy:

$n=1$: We have $\alpha (1)=\beta (1)=1$. $n=6$: We have $\alpha (6)=\frac{1+2+3+6}{4}=3$ and $\beta (6)=\frac{1+5}{2}=3$.

We conclude that $\alpha (n)=\beta (n)$ if and only if $n=1$ or $n=6$. $\square$