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# Number Theory - Italy MO 2019, Problem 2

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# Understand the problem

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Let $p,q$ be prime numbers$.$ Prove that if $p+q^2$ is a perfect square$,$ then $p^2+q^n$ is not a perfect square for any positive integer $n.$
[/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"]Italy MO 2019, Problem 2 [/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 of 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"]

# Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]Let us first write the condition and use the idea that p and q are prime. Let $a \in \mathbb{N}$ such that $$p + q^2 = a^2$$ =>   $$p = a^2 - q^2 = (a - q)(a + q)$$. As $p$ is a prime and $a + q > a - q$, then $a - q = 1$. We must have $p = 2q + 1$. Now, try to use this condition to rewrite the expression we are interested in. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]p = 2q+1. Suppose that there exists $b \in \mathbb{N}$ such that for some positive integer $n$, then
$$p^2 + q^n = b^2$$ => $$q^n = b^2 - p^2$$ => $$q^n = (b - p)(b + p)$$. It implies that both (b-p) and (b+p) must be powers of q. Hence gcd(b-p, b+p) = (b-p). b-p | b+p => b-p | 2p = 4q + 2. As, b-p is a power of q, it implies that b-p = 1 and q can be a prime or q = 2 and b-p = 2 as gcd (b-p, 4q+2) = 1 if q is an odd prime or 2 if q = 2. Thus, we get that ( b-p = 1 ) or ( b-p =2, q = 2 ) and p = 2q+1.   [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Case 1 ( b-p = 1 ): b -p = 1,  (b - p) = 1, (b+p) = $q^n = 1 + 2p = 3 + 4q$. Now, you see that LHS will grow much more fast as q increases than the RHS.  Mathematically, $q^n \geq q^2 \geq 4q+3$ as $q \geq 5, n \geq 2$. So, the only possibility for q is q = 2, 3, 5. But it has no solution as you can see. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Case 2 ( q = 2 ):  It implies p = 2q+1 = 5. $2^n = (b-5)(b+5)$. It implies that the both b-5 and b+5 must be powers of 2.  Let $b-5 = 2^a \leq b+5 = 2^b \rightarrow 2^b - 2^a = 10 \rightarrow a = 1 and 2^(b-1) - 1 = 5$. This has no solution.

QED.

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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%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

# Similar Problems

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