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Source: 2024 Korea Summer Program Practice Test P8 (original version)

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#1 h Aug 12, 2024, 9:46 AM • 2 Y

Y by Pluto1708, YS_PARKGONG

For a positive integer $n$ , let $\tau(n)$ denote the number of positive divisors of $n$ . Determine whether there exists a positive integer triple $a, b, c$ such that there are exactly $1012$ positive integers $K$ not greater than $2024$ that satisfies the following: the equation
$\tau(x) = \tau(y) = \tau(z) = \tau(ax + by + cz) = K$ holds for some positive integers $x,y,z$ .

This post has been edited 2 times. Last edited by MNJ2357, Aug 12, 2024, 10:04 AM
Reason: Added version

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#2 h Aug 12, 2024, 11:35 AM • 3 Y

Y by Pluto1708, sami1618, hellomath010118

We claim that the answer is yes. Take $(a,b,c)=(3,7,7)$
Claim 1: $3x^2+7y^2+7z^2$ is never a perfect square.
Proof: Work modulo $7$ , if $7 \nmid x$ then as $3$ is NQR and $x^2$ is QR so product must be NQR, if $7 \mid x$ , then we get that $7(21x^2+y^2+z^2)$ is perfect square, which means $7 \mid y,z$ leading to descent contradiction!.
It is well known that $\tau$ is odd only when the input is perfect square, and now Claim 1 proves that we can't construct odd $K$ , for even $K$ ; $K=2$ the construction is $x=2, y = 2, z=11$ which gives $ax+by+cz = 97$ ; now for $K=2(\ell+1)$ take $A \mapsto A \cdot p^{\ell}$ for $A \in \{2,2,11\}$ giving all even numbers, now as number of even numbers before $2024$ are $1012$ , we are done.

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#3 h Apr 3, 2025, 3:49 AM

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We claim that $(a,b,c)=(15,7,7)$ works. We make use of the following claim to avoid odd numbers:

Claim: If $a,b,c\equiv 7\pmod{8}$ , then there are no positive integers $(u,v,w)$ for which $au^2+bv^2+cw^2$ is a perfect square.

Since all squares are $0,1,4\pmod{8}$ , we need $u^2+v^2+w^2$ to be $0,4,7\pmod{8}$ . $7$ is clearly not possible since it is the sum of three squares. However, $0$ can only be achieved by $0+0+0$ or $0+4+4$ , and $4$ can only be achieved by $0+0+4$ or $4+4+4$ . Either way, $u,v,w$ are all even, so we can trigger infinite descent.

This means that odd $K$ do not work. For $K=2$ , $(x,y,z)=(3,5,11)$ works since $ax+by+cz=157$ which is also prime. For $K=2n$ , multiply all $x,y,z$ by $239^{n-1}$ , which will multiply all four $\tau$ s by $n$ .

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