2019 AIME I Problems/Problem 14

Revision as of 21:44, 6 June 2020 by Emerald block (talk | contribs) (new solution)

Problem 14

Find the least odd prime factor of $2019^8+1$.

Solution

We know that $2019^8 \equiv -1 \pmod{p}$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we find $2019^{16} \equiv 1 \pmod{p}$.

Since $2019^{16} \equiv 1 \pmod{p}$, the order of $2019$ modulo $p$ is a positive divisor of $16$.

However, if the order of $2019$ modulo $p$ is $1, 2, 4,$ or $8,$ then $2019^8$ will be equivalent to $1 \pmod{p},$ which contradicts the given requirement that $2019^8\equiv -1\pmod{p}$.

Therefore, the order of $2019$ modulo $p$ is $16$. Because all orders modulo $p$ divide $\phi(p)$, we see that $\phi(p)$ is a multiple of $16$. As $p$ is prime, $\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1$. Therefore, $p\equiv 1 \pmod{16}$. The two smallest primes equivalent to $1 \pmod{16}$ are $17$ and $97$. As $2019^8 \not\equiv -1 \pmod{17}$ and $2019^8 \equiv -1 \pmod{97}$, the smallest possible $p$ is thus $\boxed{097}$.

Note to solution

$\phi(k)$ is the Euler Totient Function of integer $k$. $\phi(k)$ is the number of positive integers less than $k$ relatively prime to $k$. Define the numbers $k_1,k_2,k_3,\cdots,k_n$ to be the prime factors of $k$. Then, we have\[\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).\]A property of the Totient function is that, for any prime $p$, $\phi(p)=p-1$.

Euler's Totient Theorem states that\[a^{\phi(k)} \equiv 1\pmod k\]if $\gcd(a,k)=1$.

Furthermore, the order $a$ modulo $n$ for an integer $a$ relatively prime to $n$ is defined as the smallest positive integer $d$ such that $a^{d} \equiv 1\pmod n$. An important property of the order $d$ is that $d|\phi(n)$.

Solution 2 (Basic Modular Arithmetic)

In this solution, $k$ will represent an arbitrary nonnegative integer.

We will show that any potential prime $p$ must be of the form $16k+1$ through a proof by contradiction. Suppose that there exists some prime $p$ that can not be expressed in the form $16k+1$ that is a divisor of $2019^8+1$. First, note that if the prime $p$ is a divisor of $2019$, then $2019^8$ is a multiple of $p$ and $2019^8+1$ is not. Thus, $p$ is not a divisor of $2019$.

Because $2019^8+1$ is a multiple of $p$, $2019^8+1\equiv0\pmod{p}$. This means that $2019^8\equiv-1\pmod{p}$, and by raising both sides to an arbitrary odd positive integer, we have that $2019^{16k+8}\equiv-1\pmod{p}$.

Then, since the problem requires an odd prime, $p$ can be expressed as $16k+m$, where $m$ is an odd integer ranging from $3$ to $15$, inclusive. By Fermat's Little Theorem, $2019^{p-1}\equiv1\pmod{p}$, and plugging in values, we get $2019^{16k+n}\equiv1\pmod{p}$, where $n=m-1$ and is thus an even integer ranging from $2$ to $14$, inclusive.

If $n=8$, then $2019^{16k+8}\equiv1\pmod{p}$, which creates a contradiction. If $n$ is not a multiple of $8$ but is a multiple of $4$, squaring both sides of $2019^{16k+n}\equiv1\pmod{p}$ also results in the same contradictory equivalence. For all remaining $n$, raising both sides of $2019^{16k+n}\equiv1\pmod{p}$ to the $4$th power creates the same contradiction. (Note that $32k$ and $64k$ can both be expressed in the form $16k$.)

Since we have proved that no value of $n$ can work, this means that a prime must be of the form $16k+1$ in order to be a factor of $2019^8+1$. The smallest prime of this form is $17$, and testing it, we get \[2019^8+1\equiv13^8+1\equiv169^4+1\equiv(-1)^4+1\equiv1+1\equiv2\pmod{17},\] so it does not work. The next smallest prime of the required form is $97$, and testing it, we get \[2019^8+1\equiv(-18)^8+1\equiv324^4+1\equiv33^4+1\equiv1089^2+1\equiv22^2+1\equiv484+1\equiv-1+1\equiv0\pmod{97},\] so it works. Thus, the answer is $\boxed{097}$. ~emerald_block

Video Solution

On The Spot STEM:

https://youtu.be/_vHq5_5qCd8


https://youtu.be/IF88iO5keFo

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS