# Understand the problem

Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square.
[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Cyprus IMO TST 2018, Problem 1 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]7/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Challenges and Thrills of Pre College Mathematics [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]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 $m^2 = 1 + n + n^2 + n^3 + n^4$.   [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]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?     [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]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? [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]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) [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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