be a polynomial with integer coefficients. Define
and 
for 
.
such that 
, then prove that either 
or 
. [/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="off" _builder_version="3.22.4"]8 out of 10
[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="on"]Polynomials - Edward Barbeau [/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" inline_fonts="Aclonica"]
Do you really need a hint? Try it first!
[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]Do you know this lemma ,Lemma: If 
and 
, then 
.
To prove this, let 
. Then
Each bracket is divisible by 
, proving the statement.
We use the fact that the sequence 
consists of only integers.
We'll first prove that we cannot have three distinct integers 
, 
, and 
such that 
, 
, and 
(In other words, the variables cannot come in a cycle of 3). Assume that there does exist such numbers. Then we should have 
, which means 
. Similarly we can get 
, which implies equality. Ultimately, it leads to two equal variables, contradiction. In a similar manner we can prove that these variables cannot come in cycles of more than 3.
Therefore, we conclude that the variables of can only come in cycles of most two. We realize that since 
,we have a cycle 
. Since the minimal cycle has length at most 2, one of 
or 
must be equal to 0, and we are done.
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