Elementary Number Theory – Prove Sequence 16, 1156, 111556, 11115556 are Squares

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I'm 16 years old, and I'm studying for my exam maths coming this monday. In the chapter "sequences and series", there is this exercise:

Prove that a positive integer formed by $k$ times digit 1, followed by $(k-1)$
times digit 5 and ending on one 6, is the square of an integer.

I'm not a native English speaker, so my translation of the exercise might be a bit crappy. What is says is that 16, 1156, 111556, 11115556, 1111155556, etc are all squares of integers. I'm supposed to prove that. I think my main problem is that I don't see the link between these numbers and sequences.
Of course, we assume we use a decimal numeral system (= base 10)

Can anyone point me in the right direction (or simply prove it, if it is difficult to give a hint without giving the whole evidence). I think it can't be that difficult, since I'm supposed to solve it.

For sure, by using the word "integer", I mean "natural number" ($\in\mathbb{N}$)

Thanks in advance.

As TMM pointed out, the square roots are 4, 34, 334, 3334, 33334, etc…

This row is given by one of the following descriptions:

  • $t_n = t_{n-1} + 3*10^{n-1}$
  • $t_n = \lfloor\frac{1}{3}*10^{n}\rfloor + 1$
  • $t_n = t_{n-1} * 10 – 6$

But, I still don't see any progress in my evidence. A human being can see in these numbers a system and can tell it will be correct for $k$ going to $\infty$. But this isn't enough for a mathematical evidence.

Best Answer

Mark Bennet already suggested looking at the numbers as geometric series, so I'll use a slightly different approach. Instead of writing the squares like that, try writing them as follows:

$$\begin{align} 15&.999\ldots = 16 \\ 1155&.999\ldots = 1156 \\ 111555&.999\ldots = 111556 \\ \vdots\end{align}$$

These numbers can be expressed as a sum of three numbers, as follows:

$$\begin{align} 111111&.111\ldots \\ 444&.444\ldots \\ 0&.444\ldots \\ \hline 111555&.999\ldots \end{align}$$

Since $1/9 = 0.111\ldots$, we get

$$\begin{align} 111111&.111\ldots = \frac{1}{9} \cdot 10^{2k} \\ 444&.444\ldots = \frac{1}{9} \cdot 4 \cdot 10^k \\ 0&.444\ldots = \frac{1}{9} \cdot 4 \\ \hline 111555&.999\ldots = \frac{1}{9} \left(10^{2k} + 4 \cdot 10^k + 4\right). \end{align}$$

But this can be written as a square:

$$\frac{1}{9} \left(10^{2k} + 4 \cdot 10^k + 4\right) = \left(\frac{10^k + 2}{3}\right)^2.$$

Since $10^k + 2$ is always divisible by $3$, this is indeed the square of an integer.

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