of the equation![\\[ (a+b+3)^2 + 2ab = 3ab(a+2)(b+2)\\]](https://latex.artofproblemsolving.com/d/e/1/de172272f7768ba93178467455664a4fd8e6af45.png)
[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][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" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Costa Rica NMO 2010[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]A Friendly Introduction to Number Theory by J.H.Silverman[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]
The idea is that the RHS is a 4 degree polynomial in two variables and the LHS is a 2 degree polynomial in the same two variables.Now, you have the intuition that the 4th-degree polynomial will surpass the LHS after some time. We, therefore, aim to bound the a and b assuming the equality holds.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]So, we try to find when RHS $latex \geq $ LHS. Rather we will find the condition when $latex (a-b)^2 \leq (ab-1)(ab + 2a + 2b + 3) $.Assume $latex a \geq b $.We will find when $latex (a-b) \leq (ab-1)$ and $latex (a-b) \leq (ab + 2a + 2b + 3)$.
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"] $latex (a-b) \leq (ab-1) $ - For this to hold.$latex (a^2-1)(b^2 -1) \geq 0$.$latex (a-b) \leq (ab + 2a + 2b + 3)$ - For this to hold$latex (a+3)(b+1) \geq 0$.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]Hence, try to understand that we have essentially bounded the solutions.Observe this implies that to solutions to have essentially.$latex -3 \leq a,b \leq 1 $.Hence show that by computation that :The solution set is 
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