  How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]Find all integer solutions $(a,b)$ of the equation $\$(a+b+3)^2 + 2ab = 3ab(a+2)(b+2)\$$

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Write the equation in this form : $(a-b)^2 = 3(ab-1)(ab+2a+2b+3)$ 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 $\geq$ LHS.  Rather we will find the condition when $(a-b)^2 \leq (ab-1)(ab + 2a + 2b + 3)$. Assume $a \geq b$. We will find when $latex (a-b) \leq (ab-1)$ and $(a-b) \leq (ab + 2a + 2b + 3)$.

[/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. $(a^2-1)(b^2 -1) \geq 0$. $(a-b) \leq (ab + 2a + 2b + 3)$ - For this to hold $(a+3)(b+1) \geq 0$. [/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 $\boxed{(-3,-3), (-3,0), (0,-3), (-2,1), (1,-2), (-1,-1), (1,1)}$ [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.26.4" 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="3" _address="0.1.0.3"]

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