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# Understand the problem

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"]

[/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" hover_enabled="0"][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="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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