$9.3$ A natural number is called square-free, if it is not divisible by the square of any prime number. For a natural number $a$, we consider the number $f(a) = a^{a+1} + 1$. Prove that: a) if $a$ is even, then $f(a)$ is not square-free b) there exist infinitely many odd $a$ for which $f(a)$ is not square-free $9.4$ We will call a generalized $2n$-parallelogram a convex polygon with $2n$ sides, so that, traversed consecutively, the $k$th side is parallel and equal to the $(n+k)$th side for $k=1, 2, ... , n$. In a rectangular coordinate system, a generalized parallelogram is given with $50$ vertices, each with integer coordinates. Prove that its area is at least $300$.
Problem
Source: Bulgarian Autumn Math Competition 2024
Tags: number theory, lattice, squarefree
Laan
21.11.2024 18:25
I have a solution for problem 9.3 $a)$ Consider $p$ is a prime factor of $a+1$. As $p|a+1$ and $a+1$ is an odd number, by LTE lenma, we have: $v_p(a^{a+1}+1)=v_p(a+1)+v_p(a+1)=2.v_p(a+1) \ge 2$(cause $p|a+1$) Therefore: $p^2|a^{a+1}+1$ $b)$For wich $a=100k+57$,k is a natural number: $f(a)=(100k+57)^{2(50k+29)}+1\equiv7^{2(50k+29)} +1\equiv (-1)+1\equiv0$(mod$25$) Hence, there exist ifinitely many odd number a s.t $f(a)$ is not square-free.
PlayForFUn
22.11.2024 00:50
Laan wrote: I have a solution for problem 9.3 $a)$ Consider $p$ is a prime factor of $a+1$. As $p|a+1$ and $a+1$ is an odd number, by LTE lenma, we have: $v_p(a^{a+1}+1)=v_p(a+1)+v_p(a+1)=2.v_p(a+1) \ge 2$(cause $p|a+1$) Therefore: $p^2|a^{a+1}+1$ $b)$For wich $a=100k+57$,k is a natural number: $f(a)=(100k+57)^{2(50k+29)}+1\equiv7^{2(50k+29)} +1\equiv (-1)+1\equiv0$(mod$25$) Hence, there exist ifinitely many odd number a s.t $f(a)$ is not square-free. What is the motivation behind looking at mod 25?
Laan
22.11.2024 14:36
PlayForFUn wrote: Laan wrote: I have a solution for problem 9.3 $a)$ Consider $p$ is a prime factor of $a+1$. As $p|a+1$ and $a+1$ is an odd number, by LTE lenma, we have: $v_p(a^{a+1}+1)=v_p(a+1)+v_p(a+1)=2.v_p(a+1) \ge 2$(cause $p|a+1$) Therefore: $p^2|a^{a+1}+1$ $b)$For wich $a=100k+57$,k is a natural number: $f(a)=(100k+57)^{2(50k+29)}+1\equiv7^{2(50k+29)} +1\equiv (-1)+1\equiv0$(mod$25$) Hence, there exist ifinitely many odd number a s.t $f(a)$ is not square-free. What is the motivation behind looking at mod 25? You can see my thinking process. $a$ is an odd number so there exists a natural number$q$ s.t: $a=2q+1$. I get:$ f(2q+1)=(2q+1)^{2.(q+1)}+1=(4q^2+4q+1)^{q+1}+1$ Here I asume $q+1$ is odd then: $(4q^2+4q+1)+1| f(2q+1)$ At this point, I search for which $q$ that $4q^2+4q+2$ has a square factor. And I see that if $q \equiv 3$(mod$25$) then $25|4q^2+4q+2$. You can choose other modulo if you want.