Question 1209775
Let's analyze each equation:

a = √(4 + √(5 + a))

Square both sides: a² = 4 + √(5 + a)
Rearrange: a² - 4 = √(5 + a)
Square both sides again: (a² - 4)² = 5 + a
Expand: a⁴ - 8a² + 16 = 5 + a
Rearrange: a⁴ - 8a² - a + 11 = 0
b = √(4 - √(7 + b))

Square both sides: b² = 4 - √(7 + b)
Rearrange: 4 - b² = √(7 + b)
Square both sides again: (4 - b²)² = 7 + b
Expand: 16 - 8b² + b⁴ = 7 + b
Rearrange: b⁴ - 8b² - b + 9 = 0
c = √(4 + √(9 - c))

Square both sides: c² = 4 + √(9 - c)
Rearrange: c² - 4 = √(9 - c)
Square both sides again: (c² - 4)² = 9 - c
Expand: c⁴ - 8c² + 16 = 9 - c
Rearrange: c⁴ - 8c² + c + 7 = 0
d = √(4 - √(11 - d))

Square both sides: d² = 4 - √(11 - d)
Rearrange: 4 - d² = √(11 - d)
Square both sides again: (4 - d²)² = 11 - d
Expand: 16 - 8d² + d⁴ = 11 - d
Rearrange: d⁴ - 8d² + d + 5 = 0
Let's look at the polynomials we derived:

P(x) = x⁴ - 8x² - x + 11 = 0 has root a
Q(x) = x⁴ - 8x² - x + 9 = 0 has root b
R(x) = x⁴ - 8x² + x + 7 = 0 has root c
S(x) = x⁴ - 8x² + x + 5 = 0 has root d
Notice the pattern. Let's consider the polynomial F(x, y) = x⁴ - 8x² + yx + (13 - 2y).

F(a, -1) = a⁴ - 8a² - a + 11 = 0
F(b, -1) = b⁴ - 8b² - b + 9 = 0
F(c, 1) = c⁴ - 8c² + c + 7 = 0
F(d, 1) = d⁴ - 8d² + d + 5 = 0
We are looking for abcd. Let's rewrite the polynomials:

a⁴ - 8a² - a + 11 = 0
b⁴ - 8b² - b + 9 = 0
c⁴ - 8c² + c + 7 = 0
d⁴ - 8d² + d + 5 = 0
Let's subtract consecutive polynomials.
Q(x) - P(x) = -2
R(x) - Q(x) = 2x - 2
S(x) - R(x) = -2x - 2
Consider the polynomials P(x), Q(x), R(x), and S(x).
Let P(x) = x⁴ - 8x² - x + 11.
Let Q(x) = x⁴ - 8x² - x + 9.
Let R(x) = x⁴ - 8x² + x + 7.
Let S(x) = x⁴ - 8x² + x + 5.
We are looking for abcd.
We can write the polynomial as:
(x-a)(x-b)(x-c)(x-d) = x⁴ - (a+b+c+d)x³ + ... + abcd.

Let's use Vieta's formulas.

a⁴ - 8a² - a + 11 = 0
b⁴ - 8b² - b + 9 = 0
c⁴ - 8c² + c + 7 = 0
d⁴ - 8d² + d + 5 = 0
Consider the polynomial (x-a)(x-b)(x-c)(x-d).
We can note that the polynomials are of the form:
x⁴ - 8x² ± x + k = 0.
Let's define F(x, y) = x⁴ - 8x² + yx + (13 - 2y) = 0.
Then a, b, c, d are roots of F(x, -1) and F(x, 1) such that a,b are roots of F(x,-1) and c,d are roots of F(x,1).
Consider the polynomial (x²-4)² - (5+x) = 0 for a.
(x²-4)² - (7+x) = 0 for b.
(x²-4)² - (9-x) = 0 for c.
(x²-4)² - (11-x) = 0 for d.
Let P(x) = (x²-4)² = x⁴ - 8x² + 16
a: P(a) - 5 - a = 0
b: P(b) - 7 - b = 0
c: P(c) - 9 + c = 0
d: P(d) - 11 + d = 0

We are looking for abcd.
We have:
(a²-4)² = a+5
(b²-4)² = b+7
(c²-4)² = 9-c
(d²-4)² = 11-d

Consider the polynomials:
P(x) = x⁴ - 8x² - x + 11 = 0
Q(x) = x⁴ - 8x² - x + 9 = 0
R(x) = x⁴ - 8x² + x + 7 = 0
S(x) = x⁴ - 8x² + x + 5 = 0

Let's look for a pattern.
Let f(k) = x^4 - 8x^2 + (-1)^k x + 13 - 2(-1)^k = 0.
k=0: d
k=1: c
k=2: b
k=3: a
The constant term of the polynomial is the product of the roots.
abcd = (13 - 2(-1)^3) = 11
abcd = (13 - 2(-1)^2) = 11
abcd = (13 - 2(-1)^1) = 11
abcd = (13 - 2(-1)^0) = 11
abcd = 11

Final Answer: The final answer is  
11
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