Question

If $2n+1$ it is a prime number, show that ${{1}^{2}},{{2}^{2}},{{3}^{2}},........,{{n}^{2}}$ when divided by $2n+1$leave different remainder.

Answer 1

Hint: Take the number $P$and $Q$so that $P=(2n+1)k+r$ and $Q=(2n+1){{k}_{1}}+r$. Now square it and subtract you will get ${{k}_{0}}$which is $\dfrac{(2n+1)}{{{P}^{2}}-{{Q}^{2}}}$. In short use a method of contradiction. So split it you will get the answer.

 Complete step-by-step answer:

A prime number (or a prime) is a natural number greater than $1$ that is not a product of two smaller natural numbers. A natural number greater than $1$ that is not prime is called a composite number. For example, $5$is prime because the only way of writing it as a product, $1\times 5$or $5\times 1$, involves$5$ itself. However, $4$ is composite because it is a product $(2\times 2)$ in which both numbers are smaller than $4$. Primes are central in number theory because of the fundamental theorem of arithmetic: every natural number greater than $1$ is either a prime itself or can be factorized as a product of primes that is unique up to their order.

The property of being prime is called primality. A simple but slow method of checking the primality of a given number $n$, called trial division, tests whether $n$ is a multiple of any integer between $2$ and $\sqrt{n}$.

A prime number is a positive integer that has exactly two positive integer factors, $1$and itself.

Note that the definition of a prime number doesn't allow $1$to be a prime number: $1$only has one factor, namely $1$. Prime numbers have exactly two factors, not "at most two" or anything like that.


When a number has more than two factors it is called a composite number.


Let $P,Q\in \left\{ 1,2,3,.......,n \right\}$,

So,

$\begin{align}

  & P=(2n+1)k+r \\

 & Q=(2n+1){{k}_{1}}+r \\

\end{align}$

So now squaring $P$and $Q$ and subtracting,

${{P}^{2}}={{(2n+1)}^{2}}{{k}^{2}}+{{r}^{2}}+2(2n+1)kr$

${{Q}^{2}}={{(2n+1)}^{2}}{{k}_{1}}^{2}+{{r}^{2}}+2(2n+1){{k}_{1}}r$

$\begin{align}

  &

{{P}^{2}}-{{Q}^{2}}={{(2n+1)}^{2}}{{k}^{2}}+{{r}^{2}}+2(2n+1)kr-{{(2n+1)}^{2}}{{k}_{1}}^{2}-{{r}^

{2}}-2(2n+1){{k}_{1}}r \\

 &

{{P}^{2}}-{{Q}^{2}}=(2n+1)((2n+1){{k}^{2}}+2kr-(2n+1){{k}_{1}}^{2}-2{{k}_{1}}r)=(2n+1){{k}_{0}}

\\

\end{align}$

i.e${{k}_{0}}=\dfrac{(2n+1)}{{{P}^{2}}-{{Q}^{2}}}$.

But ${{P}^{2}}-{{Q}^{2}}=(P+Q)(P-Q)$each factor being less than $2n$.


$\dfrac{(2n+1)}{(P+Q)(P-Q)}$, As $(2n+1)$is a prime number.

So $\dfrac{(2n+1)}{(P+Q)}$or $\dfrac{(2n+1)}{(P-Q)}$which is wrong.

The assertion that $P$and $Q$ leaves the same remainder with respect to division by

$(2n+1)$is false.

Contrary to our assumption ${{P}^{2}}$and ${{Q}^{2}}$ shall leave different remainder when

divided by $(2n+1)$.


Note: Read the question carefully. Here you must know that we have taken a number $P$and $Q$so that$P=(2n+1)k+r$ and $Q=(2n+1){{k}_{1}}+r$. So we had primarily used the method of contradiction.

Don’t confuse yourself while solving or squaring. Take the utmost care that no terms should be missing.