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Difference between revisions of "2023 IMO Problems/Problem 1"

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If <math>j \geq 2</math>, then <math>p_1^{j-1} \nmid p_1^j + p_2</math>, since the LHS has a factor of <math>p_1</math> and the RHS does not since.
 
If <math>j \geq 2</math>, then <math>p_1^{j-1} \nmid p_1^j + p_2</math>, since the LHS has a factor of <math>p_1</math> and the RHS does not since.
  
−
If <math>j = 1</math>, then the largest three factors are <math>(\frac{n}{p_2}, \frac{n}{p_1}, n)</math>. But <math>\frac{n}{p_2}</math> and <math>n</math> have a factor of <math>p_1^{e_1}</math>, while \frac{n}{p_1}<math> only has a factor of </math>p_1^{e_1 - 1}<math>, so </math>\frac{n}{p_2} \nmid \frac{n}{p_1} + n$.
+
If <math>j = 1</math>, then the largest three factors are <math>(\frac{n}{p_2}, \frac{n}{p_1}, n)</math>. But <math>\frac{n}{p_2}</math> and <math>n</math> have a factor of <math>p_1^{e_1}</math>, while <math>\frac{n}{p_1}</math> only has a factor of <math>p_1^{e_1 - 1}</math>, so <math>\frac{n}{p_2} \nmid \frac{n}{p_1} + n</math>.
  
 
Hence the prime powers are the only composite numbers which satisfy the property.
 
Hence the prime powers are the only composite numbers which satisfy the property.

Revision as of 11:44, 16 December 2023

Problem

Determine all composite integers $n>1$ that satisfy the following property: if $d_1,d_2,\dots,d_k$ are all the positive divisors of $n$ with $1=d_1<d_2<\dots<d_k=n$, then $d_i$ divides $d_{i+1}+d_{i+2}$ for every $1\le i \le k-2$.

Video Solution

https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]


https://youtu.be/FBwvqDUEDFc?si=DYrxR9wI1YaKvDm- [Video contains IMO 2023 P1 motivation + discussion] (~little-fermat)

Solution 1

If $n$ has at least $2$ prime divisors, WLOG let $p<q$ be the smallest two of these primes. Then the ordered tuple of divisors is of the form $(1,\, p,\, p^2 \dots,\, p^a,\, q \dots,\, n)$ for some integer $a\geq 1$.

To prove this claim, note that $p$ is the smallest prime that divides $n$, so it is the smallest divisor not equal to $1$, meaning the first $2$ divisors are $1$ and $p$. Furthermore, the smallest divisor of $n$ that is not equal to a power of $p$ (i.e. not equal to $(1,\, p,\, p^2\dots)$ is equal to $q$. This is because all other divisors either include a prime $z$ different from both $q$ and $p$, which is larger than $q$ (since $q$ and $p$ are the smallest two prime divisors of $n$), or don’t include a different prime $z$. In the first case, since $z>q$, the divisor is larger than $q$. In the second case, all divisors divisible by $q^2$ are also larger than $q$, and otherwise are of the form $p^x \cdot q^1$ or $p^x$ for some nonnegative integer $x$. If the divisor is of the form $p^x$, then it is a power of $p$. If it is of the form $p^x \cdot q^1$, the smallest of these factors is $p^0 \cdot q^1 = q$. Therefore, (in the case where $2$ or more primes divide $n$) the ordered tuple of divisors is of the form $(1,\, p,\, p^2 \dots,\, p^a,\, q \dots,\, n)$ for some integer $a\geq 1$, since after each divisor $p^x$, the next smallest divisor is either $p^{x+1}$ or simply $q$.

If $a\geq 2$, the condition fails. This is because $p^{a-1} \nmid p^a + q$, since $p^a$ is divisible by $p^{a-1}$, but $q$ is not since it is a prime different from $p$. If $a=1$, then $p^{a-1}=p^0=1$, which does divide $q$. Therefore $a$ must equal $1$ for the condition to be satisfied in this case. However, we know that the ordered list of divisors satisfies $d_i \cdot d_{k+1-i}=n$, meaning since the first $3$ divisors are $(1, p, q)$, then the last $3$ divisors are $(\frac{n}{q}, \frac{n}{p}, n)$, so $(\frac{n}{q})$ must divide $(\frac{n}{p} + n)$. But $\frac{n}{q}$ is divisible by $p$, so $\frac{n}{p} + n$ must also be divisible by $p$, but since $a=1$ $\frac{n}{p}$ is and $n$ isn't.

When $n=p^x$, it is easy to verify this works for all primes $p$ and all $x\geq 2$, since $p^y \vert p^{y+1} + p^{y+2}$, and the divisors are ordered as ${1,\, p,\, p^2…\, p^x}$.

~SpencerD.

Solution 2

Similar argument as above but restated.

Let $n = p_1^{e_1} p_2^{d_2} \dots p_k^{e_k}$, with $p_1 < p_2 < \dots < p_k$ primes.

If $k=1$, it's easy to verify that this property holds, since $p^l \mid p^{l+1} + p^{l+2}$.

Suppose $k \geq 2$. All divisors of $n$ which contain a factor of $p_k$ for $k \geq 2$ is at least as large as $p_2$. So, the only divisors which are possibly less than $p_2$ are of the form $p_1^{j}$. Also, by construction $p_1$ is less than $p_2$. Putting it all together, this says that the smallest factors are $(1, p_1, \dots, p_1^j, p_2)$ for some $j \geq 1$.

If $j \geq 2$, then $p_1^{j-1} \nmid p_1^j + p_2$, since the LHS has a factor of $p_1$ and the RHS does not since.

If $j = 1$, then the largest three factors are $(\frac{n}{p_2}, \frac{n}{p_1}, n)$. But $\frac{n}{p_2}$ and $n$ have a factor of $p_1^{e_1}$, while $\frac{n}{p_1}$ only has a factor of $p_1^{e_1 - 1}$, so $\frac{n}{p_2} \nmid \frac{n}{p_1} + n$.

Hence the prime powers are the only composite numbers which satisfy the property.

See Also

2023 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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