August 11, 2019

Bangladesh MO 2019 Problem 1 - Number Theory

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Understand the problem

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Find all prime numbers such that the square of the prime number can be written as the sum of cubes of two positive integers.
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Bangladesh MO 2019 Problem 1
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.26.6" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.6" open="off"]5/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.26.6" open="on"]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"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.26.6"]Write the problem in a Mathematical Equation form i.e. p^2 = a^3 + b^3. Now can you like factorize the stuff to make life easier and use divisibility rules? [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.26.6"]After factorizing, we get  p^2 = (a+b)(a^2 + b^2 - ab) . Now can use the prime factorization idea and see what are the cases possible. Observe that three cases are possible:
  • a+b = p, a^2 + b^2 - ab =p
  • a+b =p^2 , a^2 + b^2 - ab = 1
  • a+b = 1,  a^2 + b^2 - ab = p^2
Now, can you decode these cases and solve the problem like Sherlock? [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.26.6"]Observe that a, b are both positive integers. Hence the case: a+b = 1,  a^2 + b^2 - ab = p^2 is absurd. Let's concentrate on the other cases one by one. a+b =p^2 , a^2 + b^2 - ab = 1 Now,observe this that a^2 + b^2 - ab = (a-b)^2 + ab = 1 , which is has a solution iff a = b = 1. What about the other case? a+b = p , a^2 + b^2 - ab = p [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.26.6"]a+b = p, a^2 + b^2 - ab =p Observe a = - b (mod p ) this together with the second equation gives
3a^2 = 0 (modp). Now p can be 3. For p = 3, Observe that a = 1 and b = 2 is a solution. Now if p is not 3, then p must divide a and b. This implies a + b must be greater than equal to 2p, hence contradiction.

Hence the solutions are a = 1, b =1, p = 2 

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Watch video

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Connected Program at Cheenta

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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://www.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"][/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"]

Similar Problems

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