Problem 1243

In some cases, we can easily verify the statement by squaring (or even just multiplying), halving the result, and taking the square root of each part. For example:

$$\begin{aligned} & 41^{2}=\quad 1681, \quad 16=4^{2}, \quad 81=9^{2} \\ & 4901^{2}=\quad 24019801, \quad 2401=49^{2}, \quad 9801=99^{2}, \\ & 499001^{2}=249001998001, \quad 249001=499^{2}, \quad 998001=999^{2} . \end{aligned}$$

Thus, the statement of the problem is correct in those cases where we insert one or two 9s and 0s. From these cases, we conjecture that the two parts of the square of the numbers in question are not only square numbers (which is one of the statements of the problem), but the first part is precisely the square of the first half of the numbers in question, and the second part is the square of the number consisting of one more 9 than the number of 9s we inserted.

Is this always the case?

Imagine that we insert $n$ nines and then $n$ zeros between 4 and 1, where $n$ is any natural number. Thus,

$$\underbrace{49 \ldots 99}_{n+1 \text { digits }} \underbrace{0 \ldots 01^{2}}_{n+1 \text { digits }}=(\underbrace{49 \ldots 9}_{n+1 \text { digits }} \underbrace{0 \ldots 0}_{n+1 \text { digits }}+1)^{2}=\underbrace{49 \ldots+1 \text { digit }}_{\text {one }} \underbrace{0 \ldots 0^{2}}_{n+1 \text { digits } n+1 \text { digits }}+2 \cdot \underbrace{49 \ldots 9} \underbrace{0 \ldots 0}+1$$

From the first term of the sum, we can see that it ends with $2n+2$ zeros, and there are $2n+2$ digits before them, which, when read together, give the square of $49 \ldots 9$. We need to show that the remaining two-term sum is equal to $\left(10^{n+1}-1\right)^{2}$. This will also imply that this number is only $2n+2$ digits long (since the first $2n+3$ digit number is the square of $10^{n+1}$, which is $10^{2n+2}$), so it does not extend into the number before it. To solve the problem, we only need to prove that

$$\left(10^{n+1}-1\right)^{2}=2 \cdot \underbrace{49 \ldots 9}_{n+1 \text { digits }} \underbrace{0 \ldots 0}_{n+1 \text { digits }}+1$$

The number on the right can be written as:

$$2\left(5 \cdot 10^{n}-1\right) \cdot 10^{n+1}+1$$

or

$$10 \cdot 10^{n} \cdot 10^{n+1}-2 \cdot 10^{n+1}+1$$

which is

$$10^{2 n+2}-2 \cdot 10^{n+1}+1$$

This is indeed equal to the number on the left.

Remark. We could have also completed the proof by immediately writing the square of the number obtained by inserting $n$ nines and $n$ zeros in algebraic form:

$$\left[\left(5 \cdot 10^{n}-1\right) \cdot 10^{n+1}+1\right]^{2}$$

and performing the squaring:

$$\left(5 \cdot 10^{n}-1\right)^{2} \cdot 10^{2 n+2}+2\left(5 \cdot 10^{n}-1\right) 10^{n+1}+1$$

From the first term of the sum, we determine that it is a square number, the square of $\left(5 \cdot 10^{n}-1\right) \cdot 10 n+1$, and it ends with $2n+2$ zeros. For the sum of the other two terms, we determine - as before - that it is also a square number, the square of $10^{n+1}-1$, and since it is $2n+2$ digits long, it does not extend into the square number before it, its digits only add up with the zeros. However, the generality of the proof was not compromised in the form mentioned earlier, and the numbers written in the decimal system can be better visualized with fewer algebraic symbols.