This is a mathematical challenge. Prove that the sum
sqrt(53 + 20*sqrt(7)) + sqrt(284 – 64*sqrt(7))
is an integer.
Cinelli
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Integer

 Lemon Quarter
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Re: Integer
cinelli wrote:This is a mathematical challenge. Prove that the sum
sqrt(53 + 20*sqrt(7)) + sqrt(284 – 64*sqrt(7))
is an integer.
Spoiler...
In general, (x +/ y*sqrt(7))^2 = (x^2+7*y^2) +/ (2*x*y) * sqrt(7). So if we're given an expression of the form sqrt(m +/ n*sqrt(7)), we can check whether we can reduce it to the form x +/ y*sqrt(7) by seeing whether there is a solution to the pair of equations m = x^2+7*y^2, n = 2*x*y. And we can do that by determining the prime factorisation of n/2, using that to find all the possible ways of expressing n/2 as the product of two integers x and y, then calculating x^2+7*y^2 for each of those possibilities and seeing whether any of them turn out to be m.
Trying this for sqrt(53 + 20*sqrt(7)), the prime factorisation of 20/2 = 10 is 2*5, and the four possibilities for x and y are:
x y=10/x x^2+7*y^2

2^0*5^0 = 1 10 701
2^1*5^0 = 2 5 179
2^0*5^1 = 5 2 53
2^1*5^1 = 10 1 107
Since 53 is among the possibilities for x^2+7*y^2, we can read off that sqrt(53 + 20*sqrt(7)) = 5 + 2*sqrt(7).
Trying it for sqrt(284 – 64*sqrt(7)), the prime factorisation of 64/2 = 32 is 2^5, and the six possibilities for x and y are:
x y=32/x x^2+7*y^2

2^0 = 1 32 7169
2^1 = 2 16 1796
2^2 = 4 8 464
2^3 = 8 4 176
2^4 = 16 2 284
2^5 = 32 1 1031
and 284 is among the possibilities for x^2+7*y^2, allowing us to read off that sqrt(284 – 64*sqrt(7)) = 16  2*sqrt(7).
So sqrt(53 + 20*sqrt(7)) + sqrt(284 – 64*sqrt(7)) = (5 + 2*sqrt(7)) + (16  2*sqrt(7)) = 21.
Gengulphus

 Lemon Slice
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 Joined: November 9th, 2016, 11:33 am
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Re: Integer
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).
Cinelli
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).
Cinelli

 Lemon Quarter
 Posts: 3666
 Joined: November 4th, 2016, 1:17 am
 Been thanked: 2056 times
Re: Integer
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(Tt), 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 + (2ts^2)x^2 + t^2. That requires c = 2ts^2 and e = t^2, which is easily solved: the latter implies that t = +/sqrt(e), and then the former implies that s^2 = 2tc, so s = +/sqrt(2tc). 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

 Lemon Slice
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Re: Integer
Thanks very much, Gengulphus, for this extra information. Purely by chance I have come across something similar. Very probably you already know of this but it is new to me. It is known as Sophie Germain’s identity. Sophie Germain sounds to me like a modern name but, in fact, she was French and lived from 1776 to 1831. The identity is
a^4 + 4*b^4 = (a^2 – 2*a*b + 2*b^2) * (a^2 + 2*a*b + 2*b^2)
Here is an application.
Find all positive integers a, b for which a^4 + 4*b^4 is prime.
By the identity
(a^2 – 2*a*b + 2*b^2) * (a^2 + 2*a*b + 2*b^2) = p
Rewrite as
(a^2 – 2*a*b + b^2 + b^2) * (a^2 + 2*a*b + b^2 +b^2) = p
((a – b)^2 + b^2) * ((a + b)^2 + b^2) = p
Since ((a + b)^2 + b^2) > 1, it must be that
(a – b)^2 + b^2) = 1
Hence a – b = 0 and b = 1. So a = b = 1 is the only solution.
Cinelli
a^4 + 4*b^4 = (a^2 – 2*a*b + 2*b^2) * (a^2 + 2*a*b + 2*b^2)
Here is an application.
Find all positive integers a, b for which a^4 + 4*b^4 is prime.
By the identity
(a^2 – 2*a*b + 2*b^2) * (a^2 + 2*a*b + 2*b^2) = p
Rewrite as
(a^2 – 2*a*b + b^2 + b^2) * (a^2 + 2*a*b + b^2 +b^2) = p
((a – b)^2 + b^2) * ((a + b)^2 + b^2) = p
Since ((a + b)^2 + b^2) > 1, it must be that
(a – b)^2 + b^2) = 1
Hence a – b = 0 and b = 1. So a = b = 1 is the only solution.
Cinelli
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