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

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" 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"]Suppose that for a prime number $p$ and integers $a,b,c$ the following holds:
$$6\mid p+1,\quad p\mid a+b+c,\quad p\mid a^4+b^4+c^4.$$
Prove that $p\mid a,b,c$.

[/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" hover_enabled="0" _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" _i="1" _address="0.1.0.2.1" hover_enabled="0"]Show that, $a^4+b^4+c^4\equiv 2(b^2+bc+c^2)^2$ modulo $p$. Hence $p| (b^2+bc+c^2)$. From there, try to find a relation between $b$ and $c$ modulo $p$. [/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4" _i="1" _address="0.1.0.2.1"][/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]As $b^2+bc+c^2|b^3-c^3$, we have $b^3\equiv c^3\;\text{mod}\;p$. Note that, $\frac{p+1}{3}$ is an integer and hence $b^{p+1}\equiv c^{p+1}$. Using Fermat's little theorem, conclude that $b^2\equiv c^2$. Use this along with $b^3\equiv c^3$ to conclude that either $b\equiv c$ or $b,c$ are both divisible by $p$. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]Clearly, if $p|b,c$ then we are done.  Prove that the case $b\equiv c$ can be reduced to this one. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]If $b\equiv c$, then we have $a\equiv -2b$ and $a^4\equiv -2b^4$. These together give $p|18b^4$. Using $6|p+1$, conclude that $p|b$. The proof is now concluded using hint 3. [/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"]