 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Rationality certification (Posted on 2016-10-27) Given a and b are two rational numbers, fulfilling the equality
a^3 + 4a^2*b = 4a^ 2 + b^4.

Prove that the value of sqrt(a-1) is a rational number.

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) sqrt(a) is rational | Comment 1 of 2
For sqrt(a-1) to be rational I assume the constraint a>=1 is implied.  Otherwise there would be a lot of pairs (a,b) with imaginary sqrt(a-1) .

a^3 + 4a^2*b + 4ab^2 = 4a^2 + 4ab^2 + b^4.

Factor each side:
a*(a+2b)^2 = (2a + b^2)^2

Then rearrange for the lone a:
a = [(a+2b)/(2a+b^2)]^2

From this, as long as the denominator is nonzero, rational a and b will imply that a is the square of a rational number.  IE, sqrt(a) is rational.

In the case 2a+b^2=0, then a = -(b^2)/2.  Substituting yields:
-(b^2)/2*(-(b^2)/2 + 2b)^2 = 0
b^2*(b^2-4b) = 0
b=0 or 4
Then (a,b)=(0,0) or (-8,4)
0 and -8 fail the a>=1 constraint, so no new answers from the edge case.

 Posted by Brian Smith on 2016-10-28 10:37:30 Please log in:

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