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Strange algebra question

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If

$$a=\sqrt{4+\sqrt{4+a}}\\ b=\sqrt{4+\sqrt{4-b}}\\ c=\sqrt{4-\sqrt{4-c}}\\ d=\sqrt{4-\sqrt{4+d}}$$

Then what is the value of abcd?

My friends says that it is 48... but I just can't do that

MaxWong  Aug 16, 2017
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#1
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Hi Max: I'm sure that there is a SHORTER solution for "a", but my "Mathematica 11" gives this ridiculously long answer. Sorry about that.

Solve for a:
a = sqrt(sqrt(a + 4) + 4)

a = sqrt(sqrt(a + 4) + 4) is equivalent to sqrt(sqrt(a + 4) + 4) = a:
sqrt(sqrt(a + 4) + 4) = a

Raise both sides to the power of two:
sqrt(a + 4) + 4 = a^2

Subtract 4 from both sides:
sqrt(a + 4) = a^2 - 4

Raise both sides to the power of two:
a + 4 = (a^2 - 4)^2

Expand out terms of the right hand side:
a + 4 = a^4 - 8 a^2 + 16

Subtract a^4 - 8 a^2 + 16 from both sides:
-a^4 + 8 a^2 + a - 12 = 0

The left hand side factors into a product with three terms:
-(a^2 - a - 4) (a^2 + a - 3) = 0

Multiply both sides by -1:
(a^2 - a - 4) (a^2 + a - 3) = 0

Split into two equations:
a^2 - a - 4 = 0 or a^2 + a - 3 = 0

a^2 - a = 4 or a^2 + a - 3 = 0

a^2 - a + 1/4 = 17/4 or a^2 + a - 3 = 0

Write the left hand side as a square:
(a - 1/2)^2 = 17/4 or a^2 + a - 3 = 0

Take the square root of both sides:
a - 1/2 = sqrt(17)/2 or a - 1/2 = -sqrt(17)/2 or a^2 + a - 3 = 0

a = 1/2 + sqrt(17)/2 or a - 1/2 = -sqrt(17)/2 or a^2 + a - 3 = 0

a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a^2 + a - 3 = 0

a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a^2 + a = 3

a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a^2 + a + 1/4 = 13/4

Write the left hand side as a square:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or (a + 1/2)^2 = 13/4

Take the square root of both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a + 1/2 = sqrt(13)/2 or a + 1/2 = -sqrt(13)/2

Subtract 1/2 from both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a = sqrt(13)/2 - 1/2 or a + 1/2 = -sqrt(13)/2

Subtract 1/2 from both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a = sqrt(13)/2 - 1/2 or a = -1/2 - sqrt(13)/2

a ⇒ -1/2 - sqrt(13)/2 = 1/2 (-1 - sqrt(13)) ≈ -2.30278
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt((-1/2 - (sqrt(13))/(2)) + 4) + 4) = sqrt(sqrt(7/2 - (sqrt(13))/(2)) + 4) ≈ 2.30278:
So this solution is incorrect

a ⇒ sqrt(13)/2 - 1/2 = 1/2 (sqrt(13) - 1) ≈ 1.30278
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt(((sqrt(13))/(2) - 1/2) + 4) + 4) = sqrt(sqrt(7/2 + (sqrt(13))/(2)) + 4) ≈ 2.51053:
So this solution is incorrect

a ⇒ 1/2 - sqrt(17)/2 = 1/2 (1 - sqrt(17)) ≈ -1.56155
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt((1/2 - (sqrt(17))/(2)) + 4) + 4) = sqrt(sqrt(9/2 - (sqrt(17))/(2)) + 4) ≈ 2.35829:
So this solution is incorrect

a ⇒ 1/2 + sqrt(17)/2 = 1/2 (1 + sqrt(17)) ≈ 2.56155
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt((1/2 + (sqrt(17))/(2)) + 4) + 4) = sqrt(sqrt((sqrt(17) + 9)/(2)) + 4) ≈ 2.56155:
So this solution is correct

The solution is:
Answer: | a = 1/2 + sqrt(17)/2 =2.5615528........

b = b = 1/2 (1 + sqrt(13)) = 2.30277564......

c=c = 1/2 (sqrt(17) - 1) = 1.561552813.....

d =d = 1/2 (sqrt(13) - 1) = 1.30277564........

abcd = ~ 12  The product of these 4 variables certainly does not equal 48 but 12 !!!!.

Guest Aug 16, 2017
#2
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a  = sqrt [ 4 + sqrt (4 + a) ]     square both sides

a^2  = 4 + sqrt ( 4 + a)

a^2 - 4 = sqrt (4 + a)       square both sides again

a^4 - 8a^2 + 16  = 4 + a

a^4 - 8a^2 - a + 12  = 0

There are 4 possible real solutions to this

The only correct one  is  ...    a  = [ 1 + sqrt (17 ] / 2

Likewise...setting each of the other equations up as 4th power polynomials by squaring each side twice produces

b = [ 1 + sqrt (13 ) ] / 2

c = [-1 + sqrt(17) ] / 2

d =  [ -1 + sqrt (13) ] / 2

So....abcd  =  acbd  =

[ 1 + sqrt (17)] / 2  *   [-1 + sqrt(17) ] /2  *   [ 1 + sqrt (13 ) ] / 2  * [ -1 + sqrt (13) ] / 2 =

[ -1 + 17 ] / 4    *   [ -1 + 13] / 4  =

[ 16 / 4 ] * [ 12 / 4 ]  =

4  *  3   =

12

CPhill  Aug 16, 2017
edited by CPhill  Aug 16, 2017
#3
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a  = sqrt [ 4 + sqrt (4 + a) ]     square both sides

a^2  = 4 + sqrt ( 4 + a)

a^2 - 4 = sqrt (4 + a)       square both sides again

a^4 - 8a^2 + 16  = 4 + a

a^4 - 8a^2 - a + 12  = 0

There are 4 possible real solutions to this

The only correct one  is  ...    a  = [ 1 + sqrt (17 ] / 2

Likewise...setting each of the other equations up as 4th power polynomials by squaring each side twice produces

b = [ 1 + sqrt (13 ) ] / 2

c = [-1 + sqrt(17) ] / 2

d =  [ -1 + sqrt (13) ] / 2

So....abcd  =  acbd  =

[ 1 + sqrt (17)] / 2  *   [-1 + sqrt(17) ] /2  *   [ 1 + sqrt (13 ) ] / 2  * [ -1 + sqrt (13) ] / 2 =

[ -1 + 17 ] / 4    *   [ -1 + 13] / 4  =

[ 16 / 4 ] * [ 12 / 4 ]  =

4  *  3   =

12