Rusty

Today I went to math club and it made me realize how rusty I am at AMC problems. Then again, I had been having trouble thinking all day. I tried to understand what my teacher said about this one problem but I just couldn't concentrate. If I were to make an impression of myself today, I would have done a really bad job.

Problem: Find all positive integers $n$ and $p$ if $p$ is prime and \[ n^8 - p^5 = n^2+p^2 . \]

Solution:
We first manipulate the equation to get \[ n^2(n^3+1)(n^3-1)=p^2(p^3+1) \]. Clearly there's no solution for $p=2$, assume from now that $p$ is odd. Note that the factors on the left cannot pair-wisely share any factors larger than $2$, therefore $p^2$ divide exactly one of the factors, which leads to the following three cases:

Case 1: $p^2|n^2$ or $p|n\Rightarrow n=pk$ for $k\in \mathbb {N}$. However since $p^3+1\ge n^3+1$, therefore $k=1\Rightarrow n^3-1=1\Rightarrow $no solution.

Case2: $p^2|n^3-1$, let $n^3-1=p^2k$ for $k\in \mathbb {N}$. We have \[n^2k(p^2k+2)=p^3+1\], this means $p^2k+2|p^3+1\Rightarrow p^2k+2|2p-k$. Since $k<p$, therefore $2p-k\ge p^2k+2\ge p^2+2\Rightarrow 2p-1\ge p^2+2 \Rightarrow$ no solution

Case3: $p^2|n^3+1$, let $n^3+1=p^2k$ for $k\in \mathbb {N}$. We have \[n^2k(p^2k-2)=p^3+1\], this means $k<p+1,p^2k-2|p^3+1\Rightarrow p^2k-2|2p+k\Rightarrow 2p+k\ge p^2k-2$ $\ge p^2-2\Rightarrow 3p+1>p^2-2\Rightarrow p=2,3$ which gives the solution $(p,n)=(3,2)$

In conclusion after covering all possible cases, $(p,n)=(3,2)$ is the only solution. $\Box$

Note to solution: In non-rigorous terms, I feel like the key to this solution is comparing sizes.

Next Challenges(I've decided to add a few more to improve my other areas, also these problem are all from the 2004 shortlist):

1.Let $k$ be a fixed integer greater than 1, and let ${m=4k^2-5}$. Show that there exist positive integers $a$ and $b$ such that the sequence $(x_n)$ defined by

$ x_0=a,\quad x_1=b,\quad x_{n+2}=x_{n+1}+x_n\quad\text{for}\quad n=0,1,2,\dots,$

has all of its terms relatively prime to $m$.

2. Let $p$ be an odd prime and $n$ a positive integer. In the coordinate plane, eight distinct points with integer coordinates lie on a circle with diameter of length $p^{n}$. Prove that there exists a triangle with vertices at three of the given points such that the squares of its side lengths are integers divisible by $p^{n+1}$

3. Given a cyclic quadrilateral $ABCD$, let $M$ be the midpoint of the side $CD$, and let $N$ be a point on the circumcircle of triangle $ABM$. Assume that the point $N$ is different from the point $M$ and satisfies $\frac{AN}{BN}=\frac{AM}{BM}$. Prove that the points $E$, $F$, $N$ are collinear, where $E=AC\cap BD$ and $F=BC\cap DA$