Prove that polynomial p(x)=ax^2+bx+c with integer coefficients that do not simultaneously equal zero does not exist, such that p(2^(1/3))=0?

2 Answers
Mar 4, 2018

See below.

Explanation:

Given

#p(x) = a x^2+b x+c# with #{a,b,c} in ZZ^3#

we have

#p(2^(1/3))=a xx 2^(2/3)+b xx 2^(1/3) + c = 0 rArr (a xx 2^(1/3)+b)2^(1/3) = -c# and consequently

#2(a xx 2^(1/3)+b)^3 = -c^3 rArr c = -2^(3n)m^3# and then

#(a xx 2^(1/3)+b)^3 = -2^(3n-1)m^3 = p# for suitable #{n, m} in ZZ^2#

but

#a xx 2^(1/3)+b = -2^((3n-1)/3)m# is irrational because

#(3n-1)/3 notin ZZ# for any #n in ZZ# and thus we conclude that

#b = -(a xx 2^(1/3)+m xx 2^((3n-1)/3))# is irrational which is an absurd.

Mar 4, 2018

See explanation...

Explanation:

Note that #2^(1/3)# is a zero of:

#x^3-2#

Suppose #2^(1/3)# is a zero of #ax^2+bx+c#, where #a, b, c in ZZ#

Note that #a != 0# since otherwise we would have #b2^(1/3)+c = 0#, making #2^(1/3)# rational.

Further note that #b != 0# since otherwise we would have #a2^(2/3)+c = 0#, making #2^(2/3)# rational.

Then #2^(1/3)# is a zero of the monic polynomial with rational coefficients.

#x^2+b/ax+c/a#

and of:

#x^3+b/ax^2+c/ax#

So it must be a zero of:

#(x^3+b/ax^2+c/ax) - (x^3-2) = b/ax^2+c/ax+2#

and thus a zero of the monic polynomial:

#x^2+c/bx+(2a)/b#

So it must also be a zero of:

#(x^2+b/ax+c/a)-(x^2+c/bx+(2a)/b) = (b/a-c/b)x+(c/a-(2a)/b)#

SInce #2^(1/3)# is irrational, this implies that:

#{ (b/a-c/b = 0), (c/a-(2a)/b = 0) :}#

and hence:

#{ ((b/a)^2-c/a = 0), ((b/a) (c/a) -2 = 0) :}#

Hence:

#(b/a)^3 = 2#

So:

#b/a = 2^(1/3)#

which is not possible since #2^(1/3)# is irrational.