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.
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) .
Add 4ab^2 to each side:
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.