[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]Understand the problem
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Let
be prime number
s
Prove that if
is a perfect squar
e
then
is not a perfect square for any positive integer
[/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
[/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 
such that ![\[ p + q^2 = a^2 \]](https://latex.artofproblemsolving.com/9/9/9/999c69ac46ba154318fbff15d0bef4b0b15caddd.png)
=> ![\[ p = a^2 - q^2 = (a - q)(a + q) \]](https://latex.artofproblemsolving.com/5/6/4/5648ca207cf0790eaf2937ebbc1fe3ffe9a90e96.png)
.
As 
is a prime and 
, then 
. We must have 
.
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 
such that for some positive integer 
, then
![\[ p^2 + q^n = b^2 \]](https://latex.artofproblemsolving.com/6/e/3/6e377bc685b08ede2393b7e2ecc15159cf36f1c2.png)
=> ![\[ q^n = b^2 - p^2 \]](https://latex.artofproblemsolving.com/8/0/c/80ca92ab9554318e6e2b096c8c41df8c989f709c.png)
=> ![\[ q^n = (b - p)(b + p) \]](https://latex.artofproblemsolving.com/4/e/f/4ef9637676f6edf387f0497fee2a0465f9faf5a8.png)
.
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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