 # Understand the problem

Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square.
##### Source of the problem
Cyprus IMO TST 2018, Problem 1
Number Theory
7/10
##### Suggested Book
Challenges and Thrills of Pre College Mathematics

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 $(2m)^2=4n^4+4n^3+4n^2+4n+4.$ 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. $(2n^2+n)^2<4n^4+4n^3+4n^2+4n+4<(2n^2+n+2)^2$ Now, guess that $(2n^2+n)^2<(2m)^2<(2n^2+n+2)^2$ So, what we get the relationship of m and n?
We get that $(2m) = (2n^2 + n + 1)$. Hence, $(2m)^2=(2n^2+n+1)^2 \Leftrightarrow 4n^4+4n^3+4n^2+4n+4=(2n^2+n+1)^2.$ Observe that, this results in a lot of cancellation of terms and we are left with: $n^2-2n-3=0.$This gives the solution (m,n) = (11, 3)

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