Question

find that 941 is a perfect square or not by prime factorisation give full method ........​

Answer 1

no it is not a perfect square

Answer 2

By long division method, we can check out whether 941 is a perfect square or not.

Long division method is attached with the answer.

Or, let another method be preferred.

If a perfect square has the ones digit 1, or if it ends in 1, then the square root of this perfect square shall end in either 1 or 9.

$\textsf{Since \ 10x+1\equiv1\pmod{10}\ \ \ \Longrightarrow\ \ \ \pm\sqrt{10x+1}\equiv\pm\sqrt{1}\pmod{10} }\\ \\ \text{where 10x+1 is the perfect square.}\\ \\ \\ \texttt{Sometimes}\ \ \pm\sqrt{10x+1}\equiv1\pmod{10}\\ \\ \texttt{or}\ \ \pm\sqrt{10x+1}\equiv -1\pmod{10}\ \ \ \Longrightarrow\ \ \ \pm\sqrt{10x+1}\equiv 9\pmod{10}$

Thus, as the square root end in either 1 or 9, first let's assume that the square root of 941 also ends in 1.  Let it be  10x + 1.  

$\begin{aligned}&(10x+1)^2=941\\ \\ \Longrightarrow\ \ &100x^2+20x+1=941\\ \\ \Longrightarrow\ \ &100x^2+20x=940\\ \\ \Longrightarrow\ \ &20(5x^2+x)=940\\ \\ \Longrightarrow\ \ &5x^2+x=47\\ \\ \Longrightarrow\ \ &5x^2+x-47=0\end{aligned}$

Solving this quadratic equation, we won't get integer solutions.

$\begin{aligned}\Longrightarrow\ \ &\left(5x+\dfrac{1+\sqrt{941}}{2}\right)\left(x+\dfrac{1-\sqrt{941}}{10}\right)=0\\ \\ \Longrightarrow\ \ &x=-\dfrac{1\pm\sqrt{941}}{10}\end{aligned}$

Oh, here the square root of 941 also seems in the solution!!!

So, let another method be shown.

$\begin{aligned}\textsf{Consider}\ \ \ &5x^2+x&=&\ \ 47\\ \\ \Longrightarrow\ \ &10x^2+2x&=&\ \ 94\end{aligned}$

Writing the LHS and RHS decimally by splitting the digits by a slash /

$\Longrightarrow\ \ \ x^2/2x=9/4$

[Note: Slash / is used here to split the digits of the integer. Keep in mind that slash is not used for division.]

From this we get that the tens digit is the square of a number and the ones digit is double that number. But the digits of 94 can be split in many ways in some instances.

$94=9/4=8/14=7/24=6/34=5/44=4/54=3/64=2/74=1/84$

94 can be written decimally as 9/4.

$x^2/2x=9/4\\ \\ \\ \Longrightarrow\ x^2=9\ \ \ \longrightarrow\ \ \ x=3\\ \\ \\ \text{But}\\ \\ \\ \Longrightarrow\ 2x=4\ \ \ \longrightarrow\ \ \ x=2$

94 can be written decimally as 4/54.

$x^2/2x=4/54\\ \\ \\ \Longrightarrow\ x^2=4\ \ \ \longrightarrow\ \ \ x=2\\ \\ \\ \text{But}\\ \\ \\ \Longrightarrow\ 2x=54\ \ \ \longrightarrow\ \ \ x=27$

94 can be written decimally as 1/84.

$x^2/2x=1/84\\ \\ \\ \Longrightarrow\ x^2=1\ \ \ \longrightarrow\ \ \ x=1\\ \\ \\ \text{But}\\ \\ \\ \Longrightarrow\ 2x=84\ \ \ \longrightarrow\ \ \ x=42$

Each three case is not possible. Especially the last two cases have no chances for being possible!

Hence found that 941 is not a square of an integer which ends in 1.

Now let's assume that the square root of 941 ends in 9.  Let it be  10x + 9.

$\begin{aligned}&(10x+9)^2=941\\ \\ \Longrightarrow\ \ &100x^2+180x+81=941\\ \\ \Longrightarrow\ \ &100x^2+180x=860\\ \\ \Longrightarrow\ \ &20(5x^2+9x)=860\\ \\ \Longrightarrow\ \ &5x^2+9x=43\\ \\ \Longrightarrow\ \ &5x^2+9x-43=0\\ \\ \Longrightarrow\ \ &\left(5x+\dfrac{9+\sqrt{941}}{2}\right)\left(x+\dfrac{9-\sqrt{941}}{10}\right)=0\\ \\ \Longrightarrow\ \ &x=-\dfrac{9\pm\sqrt{941}}{10}\end{aligned}$

Here the square root of 941 also seems in the solution.

$\begin{aligned}\textsf{Consider}\ \ \ &5x^2+9x&=&\ \ 43\\ \\ \Longrightarrow\ \ &10x^2+18x&=&\ \ 86\\ \\ \Longrightarrow\ \ &x^2/18x&=&\ \ 86\end{aligned}$

We get no unique solutions for this splitting.

Hence found that 941 is not a perfect square of an integer which ends in 9 too.

Hence Verified!!!