# Understand the problem

Determine all integers for which the number in base is a perfect square.

##### Source of the problem

Cyprus IMO TST 2018, Problem 1

##### Topic

Number Theory

##### Difficulty Level

7/10

##### Suggested Book

Challenges and Thrills of Pre College Mathematics

# Start with hints

Do you really need a hint? Try it first!

Let us write the problem in Mathematical Language i.e. in the form of equations. \( (11111)_n\) in base n \( = 1 + n + n^2 + n^3 + n^4 \). So, the problem reduces to finding positive integer solutions to \( m^2 = 1 + n + n^2 + n^3 + n^4 \).

The idea is that we will try to bound the \( 1 + n + n^2 + n^3 + n^4 \) in between some squares and from that we will try to estimate the values of m in terms of n. Observe that Now, can you form squares from the right side? If not can you bound it by two squares?

First of all to form, you take the max terms \(4n^4 = (2n)^2 \). So, that term must be included in the square. Also, try to find a, b, c such that \( (2n^2 + an + b)^2\) can be made greater or lesser the given expression. Observe that you will get the following. Now, guess that So, what we get the relationship of m and n?

We get that \((2m) = (2n^2 + n + 1) \). Hence, Observe that, this results in a lot of cancellation of terms and we are left with: This gives the solution (m,n) = (11, 3)

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