Prove that for any real values of c, the equation x ( x 2 </msup> &#x2212;<!-- −

Joshua Foley

Joshua Foley

Answered question

2022-06-29

Prove that for any real values of c, the equation x ( x 2 1 ) ( x 2 10 ) = c can't have 5 integer solutions.

Answer & Explanation

thatuglygirlyu

thatuglygirlyu

Beginner2022-06-30Added 14 answers

Here's a more "elementary" way.
Let a,b,c,d,e be integer roots. We attempt to find a contradiction.
By Vieta's formulas, we have:
a + b + c + d + e = 0
a b + a c + a d + a e + . . . + d e = 11
So
22 = ( equation 1 ) 2 2 ( equation 2 )
= a 2 + b 2 + c 2 + d 2 + e 2
Note that a 2 , b 2 , c 2 , d 2 , e 2 must each be one of 0,1,4,9,16.
One can quickly check (there are not that many cases) that, up to permutation, only ( a 2 , b 2 , c 2 , d 2 , e 2 ) = ( 16 , 4 , 1 , 1 , 0 )  or  ( 9 , 4 , 4 , 4 , 1 , 0 )
This implies that the constant of our polynomial "C" (capital) is 0.

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