# Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" box_shadow_style="preset2"]Find all pairs of prime numbers $(p, q)$ for which $7pq^2 + p = q^3 + 43p^3 + 1$ [/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" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Dutch MO 2015 Problem 4 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]5/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Challenges and Thrills in Pre College Mathematics [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="4.0" hover_enabled="0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]This Diophantine Equation may seem a bit difficult to handle and will force you to try various techniques like making modulo 7, modulo p, modulo q as p and q are given as primes. But, let's go through the basic techniques for handling it. So what is it? Checking the parity of p and q in the given equation. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]Som check that if both p and q are odd primes, then the LHS will be even but the RHS will be odd, which is a contradiction.  Hence the only way it can happen that one of them must be even i.e. 2. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Now, things seem to be under control. We have two cases, p = 2 and q = 2. For p = 2, we get the equation $q^3 - 14q^2 = -343$. This implies that q must divide $343 = 7^3$. Hence q can be only 7. This gives rise to the solution (2,7). The next hint offers the other case. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]For  q = 2. we get $43p^3 - 29p + 9 = 0$. How to solve this? Clearly apply the same idea. Observe that if p = odd the LHS will be odd which can't be 0. Hence, p must be 2, but it doesn't satisfy the equation. The only solution is (2,7). [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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