Here is a video solution for ISI Entrance Number Theory Problems based on Sum of 8 fourth powers. Watch and Learn!
How can you solve this fourth-degree diophantine equation? Let's see in the video below:
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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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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]
RMO 2018 Tamil Nadu Problem 3 is from Number Theory. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.
Show that there are infinitely many 4-tuples (a, b, c, d) of natural numbers such that $a^3 + b^4 + c^5 = d^7$.
Also see
Powers of 3 have a very interesting property: $$ 3^n + 3^n + 3^n = 3^{n+1} $$
This simple observation is the key to this problem.
We want $ a^3 = 3^n, b^4 = 3^n, c^5 = 3^n $. Clearly n needs to be a multiple of 3, 4 and 5. For example, $3 \times 4 \times 5 = 60 $ may work. That is $$ 3^{60} + 3^{60} + 3^{60} = (3^{20})^3 + (3^{15})^4 + (3^{12})^5 $$
Hence in this case $$ a = 3^{20}, b = 3^{15}, c = 3^{12} $$
This will work for any multiple of 60. Suppose 60k is a multiple of 60 (that is k is any integer). Then we will have $$ a = 3^{20k}, b = 3^{15k}, c = 3^{12k} $$
Notice that $$ 3^{60k} + 3^{60k} + 3^{60k} = (3^{20k})^3 + (3^{15k})^4 + (3^{12k})^5 = 3^{60k+1} $$
We need $$ 3^{60k+1} = d^7 $$
That is 60k +1 needs to be a multiple of 7. In terms of modular arithmetic we want $$ 60k + 1 \equiv 0 \mod 7 $$
$60 \equiv 4 \mod 7 \ \Rightarrow 60k + 1 \equiv 4k +1 \mod 7 \ \Rightarrow 4k \equiv - 1 \equiv 6 \mod 7$
This is where we will use the notion of inverse of a number modulo 7. Inverse of 4 modulo 7 is 2. This is because $ 4 \times 2 = 8 \equiv 1 \mod 7 $. The Bezout's theorem guarantees existence of inverse of 4 modulo 7. (Look into the reference at the end of this discussion if you do not know these ideas).
$4k \equiv 6 \mod 7 \ \Rightarrow 2\times 4k \equiv 2\times 6 \mod 7 \ \Rightarrow k \equiv 5 mod 7 $
Hence k = 7k' + 5 is suitable for our purpose.
Since there are infinitely many such integers with have infinitely many 4 tuples that will work.
Illustration: For k' = 0, k = 5. Therefore $60 \times 5 = 300$ should work. And it does: $$ 3^{300} + 3^{300} + 3^{300} = (3^{100})^3 + (3^{75})^4 + (3^{60})^5 = 3^{301} = (3^{43})^7 $$
Also see