cinelli wrote:Gengulphus has an elegant solution to the problem. I too tried to solve the simultaneous equations (for the first expression)
53 = x^2 + 7*y^2
20 = 2*x*y
but rather than looking for factors of x*y, I eliminated y to give an equation in x:
x^2 + 7*(10/x)^2 = 53
Multiplying through by x^2, this becomes
x^4 – 53*x^2 + 700 = 0
a quadratic in x^2 which can be factorised as
(x^2 – 25) * (x^2 – 28) = 0
As we are looking for an x which is not a multiple of sqrt(7), x = +/-5 is the solution and the corresponding y is +/-2.
I have seen another method which I include for interest. Suppose
sqrt(53 + 20*sqrt(7)) = c
Then
53 + 20*sqrt(7) = c^2 or 20*sqrt(7) = c^2 – 53
Squaring again this becomes
2800 = c^4 – 106*c^2 + 2809 or c^4 – 106*c^2 + 9 = 0
The next step is surely like pulling a rabbit out of a hat because this can be factorised as
(c^2 + 10*c -3) * (c^2 – 10*c -3) = 0
and each quadratic factor can be solved by the formula to give
c = +/-5 +/-2*sqrt(7)
Of these four values, only one is the c we started with and by observation, this is 5 + 2*sqrt(7).
The method of pulling the rabbit out of the hat might be of interest:
We've got a quartic equation Ax^4 + Bx^3 + Cx^2 + Dx + E = 0 in which the linear and cubic terms are zero, i.e. B = D = 0. Divide through by A (which is not 0, otherwise it isn't a quartic equation) to change it to:
x^4 + cx^2 + e = 0
where c = C/A and e = E/A.
We look for ways to express this as the product of two quadratics, say as (px^2 + qx + r) (Px^2 + Qx + R) = pPx^4 + (pQ+qP)x^3 + (pR+qQ+rP)x^2 + (qR+rQ)x + qQ = 0. Equating coefficients of x^4 gives us 1 = pP, so we can divide (px^2 + qx + r) through by p and multiply (Px^2 + Qx + R) through by p to express it as (x^2 + sx + t) (x^2 + Sx + T) = x^4 + (s+S)x^3 + (t+sS+T)x^2 + (tS+sT)x + tT = 0, where s = q/p, t = r/p, S=Qp, T=Rp.
Equating the coefficients of x^3 and x gives us 0 = s+S, so S = -s. Then equating the coefficients of x gives us 0 = tS+sT = -ts+sT = s(T-t), from which it follows that either s = 0 or T = t. So we can try to find an expression of x^4 + cx^2 + e as a product of two quadratics in two ways, either by setting S = -s and s = 0, i.e. s = S = 0, or by setting S = -s and T = t. The first leads to us trying to express x^4 + cx^2 + e as (x^2 + t) (x^2 + T), or equivalently to express y^2 + cy + e as (y + t) (y + T) = y^2 + (t+T)y +tT: this is just your first approach above of treating x^4 + cx^2 + e = 0 as a quadratic in y = x^2.
The second leads to us trying to express x^4 + cx^2 + e as (x^2 + sx + t) (x^2 - sx + t) = x^4 + (2t-s^2)x^2 + t^2. That requires c = 2t-s^2 and e = t^2, which is easily solved: the latter implies that t = +/-sqrt(e), and then the former implies that s^2 = 2t-c, so s = +/-sqrt(2t-c). In this case, e = 9, so t = +/-3, and c = -106, so if t = 3, s = +/-sqrt(6-(-106)) = +/-sqrt(112) = +/-4sqrt(7), and if t = -3, s = +/-sqrt((-6)-(-106)) = +/-sqrt(100) = +/-10.
So we have four solutions for the pair (s,t), namely (3,+/-4sqrt(7)) and (-3,+/-10). The first two allow us to pull one rabbit out of the hat (with the two solutions corresponding to the two possible orders of the factors):
x^4 - 106x^2 + 9 = (x^2 + 4sqrt(7)x + 3) (x^2 - 4sqrt(7)x + 3)
and the second two allow us to pull a second rabbit out of the hat (again with the two solutions corresponding to the two possible orders of the factors):
x^4 - 106x^2 + 9 = (x^2 + 10x - 3) (x^2 - 10x - 3)
In this case, the second rabbit is of particular interest due to only using integers.
Gengulphus