(PQ)²=RSP, where P,Q,R,S are distinct single digit natural numbers, then R=? I have a problem I tried to solve, but couldn't because I don't know the method to solve it and I've never come across such problem.
Here's the problem.
$(PQ)²=RSP$
Where $P, Q, R, S$ are distinct single digit natural numbers, then $R=$?
We need to find out the value of $R$, and given option are
a) $1$
b) $2$
c) $3$
d) $4$
e) $5$
I've tried solving it by option with newly created equation as $Q^2=\frac{RS}{P}$ and taking values of $R$ from above options and other variables as distinct single digit numbers other than $R$, I've gone to a level after which I've got myself a lot confused and didn't know what to do. I think there is easier method to solve it which I don't know.
I'd appreciate if someone could give me a hand with this. Thanks in advance. :)
PS: The answer is c) $3$
 A: The steps are as follows.


*

*A square number can only end in $1, 4, 5, 6, 9$.  To prove this one uses modular arithmetic.  So $P = 1, 4, 5, 6, 9$.

*$40^2 > 999$, so $P = 1$.  

*Since $(PQ)^2$ ends in $1$, it follows that $Q = 1, 9$.

*Since $P \neq Q$, it follows that $Q = 9$.

*So the digits are $19^2 = 361$.

A: Generalizing from decimal to arbitrary radix, a quick computer search seems to indicate  that there are only a handful of radices with such a unique solution. This leads to a proof of the following 
THEOREM $\;$ If in radix $\rm M$ notation we have $\rm (PQ)^2 = RSP$ with $\rm P,Q,R,S$ distinct digits and this solution is unique in radix $\:\rm M \:$ then it is one of the following, where $\;\rm A = 10,\; B = 11, \:\ldots,\: I = 18 \;$. 
$$\begin{array}{|r|r|r|} 
\hline
\rm M & \rm PQ & \rm RSP \\
\hline
6 & 15 & 321 \\
7 & 23 & 562 \\
8 & 17 & 341 \\
9 & 18 & 351 \\
10 & 19 & 361 \\
11 & \rm 1A & 371 \\
18 & \rm 1H & \rm 3E1 \\
19 & \rm 1\:I & \rm 3F1 \\
\end{array}$$
Proof: $\;$ The two general solutions listed below prove that all radices $\;\rm M > 26 \;$ have nonunique solution (except possibly $\rm M = 44\:$). The remaining small number of exceptional cases were verified by computer.   
$\quad\quad\rm  (1\: M + M-1)^2 \;=\; \;\;\: 3\: M^2 + (M\;-\;\:4) M + 1, \quad\quad M > 7$   
$\quad\quad\rm  (4\: M + M-2)^2 \;=\; 24\: M^2 + (M-20) M + 4, \quad\quad M>26,\;\; M \ne 44$ 
