# (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) $3$

• Do PQ and RSP indicate concatenation? You should say this. – Qiaochu Yuan Aug 26 '10 at 16:37
• I'm not sure. They haven't given any more information about it. But I'm presuming they mean multiplication. Like this - (PQ)²=RS*P I could be wrong though. :\ – Electrifyings Aug 26 '10 at 16:42
• Oh. You don't seem to have realized this! PQ does not mean P times Q, it means the number with first digit P and second digit Q. – Qiaochu Yuan Aug 26 '10 at 16:43
• Qiaochu: You're right, but it can be seen why he'd think multiplication is happening; joining two variables as concatenation is something you don't see very frequently. – J. M. is a poor mathematician Aug 26 '10 at 16:47
• Retagged from "numerical-methods" (which this definitely isn't) to "arithmetic." – whuber Aug 26 '10 at 18:30

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 $19^2 = 361$.
• OMG, thanks for clearing this up. Yeah, I figured this by trial and error and by going with options. LOL, all this time I was thinking its a multiplication and never had a single strike in my mind about concatenation. Thanks man for clearing it up! You made my day! :D – Electrifyings Aug 26 '10 at 16:50

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.

$\quad\quad\rm (1\: M + M-1)^2 \;=\; \;\;\: 3\: M^2 + (M\;-\;\:4) M + 1, \quad\quad M > 7$

$\quad\quad\rm (4\: M + M-2)^2 \;=\; 24\: M^2 + (M-20) M + 4, \quad\quad M>26,\;\; M \ne 44$

• Nice conjecture. But when M > 81 you can check that 93^2 = (9M+3)^2 = 81M^2 + 54M + 9 = (81)(54)9; i.e., (M,P,Q,R,S) = (M,9,3,81,54) is a solution for all M >= 82. – whuber Aug 27 '10 at 20:50
• @whuber: The conjecture is true - see the above proof. – Bill Dubuque Jan 12 '11 at 23:38