Polynomial but rearranged

Algebra Level 3

Let f ( x ) f(x) be a polynomial with integer coefficients such that f ( 0 ) = 2019 f(0)=2019 and f ( 1 ) = 9021 f(1)=9021 . Does the polynomial f ( x ) f(x) have integer roots ?

No Yes

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chris Lewis
Apr 17, 2019

Since a b a n b n a-b|a^n-b^n for distinct integers a a and b b and integer n 1 n\ge 1 , it follows that (for a b a \neq b ), a b f ( a ) f ( b ) a-b|f(a)-f(b) .

Let's say r r is an integer root of f f . Then applying the above with the given information, both

r 2019 r|2019 and r 1 9021 r-1|9021

have to hold. But checking the divisors (positive and negative) of these shows that no such r r exists.

Nice! Note that a parity argument solves a much more general case and the divisors need not be checked.

Sathvik Acharya - 2 years, 1 month ago

Log in to reply

@Sathvik Acharya , Can you please elaborate?

Mr. India - 2 years, 1 month ago

Log in to reply

Just consider everything modulo 2 2 . If both f ( 0 ) f(0) and f ( 1 ) f(1) are odd, f ( n ) f(n) must always be odd for integer n n . And zero is not an odd number...

Chris Lewis - 2 years, 1 month ago

Log in to reply

@Chris Lewis Thank you.

Mr. India - 2 years, 1 month ago

Haha, yes, that is much quicker!

Chris Lewis - 2 years, 1 month ago

@Sathvik Acharya , Can you please elaborate on what do you mean when you say you apply parity argument to the above question?

Vinayak Nayak - 2 years, 1 month ago

Log in to reply

See the replies to that comment

Mr. India - 2 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...