Purr-fection

Peter of Madras College, St Andrew's employed an exhaustive search to find the smallest perfect square that ends in 9009 and came up with the answer 1503.

I first noted that the last digit has to be a 3 or a 7 for the square to end in 9. Noting that the last two digits of $x^2$ are only affected by the last two digits of $x$. I then systematically went through all the squares.

I kept a record of the numbers tried in two tree diagrams starting from the units digits 3 and 7. If any of these produced a number that ended in 09 then I marked that as the next branch point on the diagram.

I then went on to further generations looking for numbers ending in 009, and then finally the next generation looking for numbers ending in 9009. I found that there are no numbers with 3 digits or less whose squares end in 9009 and the four digit numbers are 1503, 6503, 2753, 7753, 2247, 7247, 3497 and 8497.

Alternatively suppose $x^2 = 100a + 10b + c$ where $a$, $b$ and $c$ are whole numbers, $a \geq 1$ and $b$ and $c$ are between 0 and 9 inclusive.

$$x^2-9=(x-3)(x+3)=\star \star \star \star 9000$$

As 10 divides the right hand side of this expression we know 10 divides $x - 3$ or $x + 3$. Thus $x$ ends in a 3 or a 7.

Case 1: c = 3

$(100a + 10b + 3)^2 = 10000a^2 + 200a(10b + 3) + 100b^2 + 60b + 9$ ends in 9009 Subtract 9, then take modulo 100.

$$\begin{array}{rl}\Rightarrow 60b& \equiv 0\text{(mod 100)}\\ \Rightarrow 3b& \equiv 0\text{(mod 5)}\\ \Rightarrow b& =0or5.\end{array}$$

If c = 3 and b = 0:

$(100a + 3)^2$ ends in 9009

$$\begin{array}{rl}\Rightarrow 600a& \equiv 9000\text{(mod10000)}\\ \Rightarrow 6a& \equiv 90\text{(mod 100)}\\ \Rightarrow 3a& \equiv 45\text{(mod 50)}\\ \Rightarrow a& =15+50k\\ \Rightarrow \text{smallest}a& =15;x=1503.\end{array}$$

If c = 3 and b = 5:

$(100a + 53)^2$ ends in 9009 $10000a^2 + 10600a + 2809$ ends in 9009 $100a^2 + 106a + 28$ ends in 90

$$\begin{array}{rl}\Rightarrow 6a+28& \equiv 90\text{(mod100)}\\ \Rightarrow 3a& \equiv 31\text{(mod 50)}\\ \Rightarrow 3a& =31+50k\text{where k and a are nonnegative integers.}\\ \Rightarrow \text{smallest}a& =27;x=2753\text{which is not minimal.}\end{array}$$

Case 2: c = 7

$(100a + 10b + 7)^2 = 10000a^2 + 200a(10b + 7) + 100b^2 + 140b + 49$ ends in 9009.

$$\begin{array}{rl}\Rightarrow 40b+40& \equiv 0\text{(mod100)}\\ \Rightarrow 2b+2& \equiv 0\text{(mod 5)}\\ \Rightarrow b& =4\text{or}9\end{array}$$

In the cases $c$ = 7 and $b$ = 4 or 9 there are no solutions less than 1503.