Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If m is a prime number and $a,b$ two numbers less than $m$, prove that ${{a}^{m-2}}+{{a}^{m-3}}b+{{a}^{m-4}}{{b}^{2}}+...+{{b}^{m-2}}$ is a multiple of $m$.

Last updated date: 13th Jun 2024
Total views: 385.8k
Views today: 9.85k
Verified
385.8k+ views
Hint: We use the Fermat’s little theorem which states that ${{a}^{p-1}}\equiv 1\left( \bmod p \right)$ where $p$ is a prime and $a$ is an integer such that $p$ does not divide $a$ to prove ${{a}^{m-1}}-1,{{b}^{m-1}}-1$ are multiples of $m$. We also use the algebraic identity for some positive integer $n$ as ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-2}}b+{{a}^{n-3}}{{b}^{2}}+...+{{b}^{n-1}} \right)$

Complete step-by-step solution:
We know from Fermat’s theorem of modular arithmetic that if $p$ is a prime number and $a$ is any integer such that $a$ is not a multiple of $p$ then ${{a}^{p}}-a$ is multiple of $p$. Mathematically we have;
\begin{align} & p|{{a}^{p}}-a \\ & \Rightarrow {{a}^{p}}\equiv a\left( \bmod p \right) \\ & \Rightarrow {{a}^{p-1}}\equiv 1\left( \bmod p \right) \\ & \Rightarrow p|{{a}^{p-1}}-1 \\ \end{align}
We are given the question that $m$ is a prime number and $a,b$ two numbers less than$m$. Since $a,b$ are less than $m$ they cannot be a multiple of $m$. So b y Fermat’s little thermo we can state that ${{a}^{m-1}}-1$ and ${{b}^{m-1}}-1$ are multiple of $m$ that is
$m|{{a}^{m-1}}-1,m|{{b}^{m-1}}-1$
We know that if a number divides two numbers then the number also divides their sum or difference. So we have
\begin{align} & \Rightarrow m|{{a}^{m-1}}-1-\left( {{b}^{m-1}}-1 \right) \\ & \Rightarrow m|{{a}^{m-1}}-{{b}^{m-1}} \\ \end{align}
We use the factorization identity ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-2}}b+{{a}^{n-3}}{{b}^{2}}+...+{{b}^{n-1}} \right)$ for $n=m-1$ in the above step to have;
$\Rightarrow m|\left( a-b \right)\left( {{a}^{m-2}}+{{a}^{m-3}}{{b}^{2}}+{{a}^{m-4}}{{b}^{3}}+...+{{b}^{m-2}} \right)$
We in the above step $m$ either exactly divide $a-b$ or $m$ exactly divide ${{a}^{m-2}}+{{a}^{m-3}}{{b}^{2}}+{{a}^{m-4}}{{b}^{3}}+...+{{b}^{m-2}}$ or divide both. We are given that $a,b < m$ , so $m$ cannot exactly divide $a-b$. So it can only exactly divide ${{a}^{m-2}}+{{a}^{m-3}}{{b}^{2}}+{{a}^{m-4}}{{b}^{3}}+...+{{b}^{m-2}}$. Hence it is proved that ${{a}^{m-2}}+{{a}^{m-3}}{{b}^{2}}+{{a}^{m-4}}{{b}^{3}}+...+{{b}^{m-2}}$ is a multiple of $m$. 

Note: We note that if a divisor $d$ divides a number $n$ with remainder zero we call the division exact division, the divisor $d$ a factor of $n$ and the dividend $n$ a multiple of $d$. We know from law modular arithmetic that if $a\equiv b\left( \bmod n \right)\Leftrightarrow ak\equiv bk\left( \bmod n \right)$ for non-zero integers $k$. The converse of Fermat’s little theorem is used for primality of number if there exist integer $a,p$ such that ${{a}^{p}}\equiv p\left( \bmod n \right)$ for all primes $q$ dividing exactly $p-1$ and $p$ cannot divide ${{a}^{\dfrac{p-1}{q}}}-1$ then $p$ is a prime.