# Understand the problem

Find all prime numbers such that the square of the prime number can be written as the sum of cubes of two positive integers.
[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.26.6"]
[/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"]

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

[/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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