Tuesday, 23 May 2017

polynomials - When is sqrt[3]a+sqrtb+sqrt[3]asqrtb an integer?

I saw a Youtube video in which it was shown that
(7+501/2)1/3+(7501/2)1/3=2
Since there are multiple values we can choose for the 3rd root of a number, it would also make more sense to declare the value of this expression to be one of 2,1+6, or 16



We may examine this more generally. If we declare x such that
x=(a+b1/2)1/3+(ab1/2)1/3
(supposing a and b to be integers here)
one can show that

x3+3(ba2)1/3x2a=0
Which indeed has 3 roots.



We now ask




For what integer values of a and b is this polynomial solved by an integer?




I attempted this by assuming that n is a root of the polynomial. We then have

x3+3(ba2)1/3x2a
||
(xn)(x2+cx+d)
||
x3+(cn)x2+(dnc)xnd
Since (cn)x2=0 we conclude that c=n and we have
x3+3(ba2)1/3x2a=x3+(dc2)xcd
And - to continue our chain of conclusions - we conclude that
3(ba2)1/3=dc2and2a=cd
At this point I tried creating a single equation and got

108b=4d3+15c2d2+12c4dc6
This is as far as I went.

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