Evaluating a Big Fraction

$x^4 + 64 = (x^2 + 8)^2 - 16x^2 = (x^2 -4x + 8)(x^2+4x+8) = ( (x-2)^2+4)( (x+2)^2 + 4)$ .

So the product is equal to $\frac{(6^2 + 4)(10^2+4)(14^2+4)(18^2+4)\ldots (62^2+4)(66^2+4)}{(2^2+4)(6^2+4)(10^2+4)(14^2+4)\ldots(58^4+4)(62^2+4)}$ . Thus the product telescopes (terms cancel out) and we obtain $\frac {66^2 + 4}{2^2 + 4} = 545$ .

Notice that all factors in both numerators and denominators can actually be written in the form of $(4^4 . a^4+ 4^3)$ . For examples: $(8^4+64)=((4.2)^4+64)=(4^4.2^4+4^3).$ And also, $(60^4+64)=((4.15)^4+64)=(4^4.15^4+4^3).$

So, changing all the factors, the question can be rewritten as : $\frac {(4^4.2^4+4^3)(4^4.4^4+4^3)(4^4.6^4+4^3)...(4^4.14^4+4^3)(4^4.16^4+4^3)} {(4^4.1^4+4^3)(4^4.3^4+4^3)(4^4.5^4+4^3)...(4^4.13^4+4^3)(4^4.15^4+4^3)}$

Notice that somehow, all factors in both numerators and denominators have the factor of 64= $4^3$ (or you can say all factors are actually divisible by 64= $4^3$ .

So, we divide all factors by $4^3$ , and we get : $\frac {(4.2^4+1)(4.4^4+1)(4.6^4+1)...(4.14^4+1)(4.16^4+1)} {(4.1^4+1)(4.3^4+1)(4.5^4+1)...(4.13^4+1)(4.15^4+1)}$

The factors are actually written in form of $4a^4+1$ . The form is so uniform throughout all the factors, so it should have something special. After trying out some possibilities, we can actually know that : $(4a^4+1) = (2a^2+2a+1)(2a^2-2a+1)$

So for example, $(4.2^4+1) = (2.2^2+2.2+1)(2.2^2-2.2+1)$

We change the fraction again into those factors, and we have : $\frac {(2.2^2+2.2+1)(2.2^2-2.2+1)(2.4^2+2.4+1)(2.4^2-2.4+1)...(2.14^2+2.14+1)(2.14^2-2.14+1)(2.16^2+2.16+1)(2.16^2-2.16+1)} {(2.1^2+2.1+1)(2.1^2-2.1+1)(2.3^2+2.3+1)(2.3^2-2.3+1)...(2.13^2+2.13+1)(2.13^2-2.13+1)(2.15^2+2.15+1)(2.15^2-2.15+1)}$

Notice that factors that follow the form of $(2a^2+2a+1)$ is always on the opposite sides of the fraction of $(2(a+1)^2-2(a+1)+1)$ , like for example : with $a=1$ , the factor $(2.1^2+2.1+1)$ is on the denominator , and with $a=1, a+1=2$ , the factor $(2.2^2-2.2+1)$ is on the numerator.

And also if the place is vice versa, it still holds; when $a=2$ , the factor $(2.2^2+2.2+1)$ is on the numerator, and with $a=2, a+1=3$ , the factor $(2.3^2-2.3+1)$ is on the denominator. So then I assumed that those factors will cancel out eachother, since each of them will be opposite to another.

Here's the proof : $(2a^2+2a+1)=(2(a+1)^2-2(a+1)+1)$

$2a^2+2a+1=2(a^2+2a+1)-2a-2+1$

$2a^2+2a+1=2a^2 + 4a + 2 -2a -2 +1$

$2a^2+2a+1=2a^2+2a+1$

So, after the cancellations of the same values of factors on both numerator and denominator are looked after, the remaining factor on the numerator should be the $(2a^2+2a+1)$ factor, with highest value of $a$ possible, because the factor to cancel it must have higher $a$ value, which is not exist in the question. So it should be $a=16$ and the factor is : $(2.16^2+2.16+1)=545$ .

While the factor left on the denominator should be $(2(a+1)^2-2(a+1)+1)$ factor with the lowest value of $(a+1)$ possible, because the factor to cancel it must have lower value of $a$ , which is not exist in the question. So it should be $(a+1)=1$ and the factor is : $(2.1^2-2.1+1)=1$ .

So, the final fraction after the cancellation would be : $\frac {545} 1$ At last, the final answer is $545$ .

We first factor out $2^6$ from each of the terms. This leaves us with

$\frac {\prod_{i=1}^{8} 4 \cdot (2i)^4 +1} {\prod_{i=1}^{8} 4 \cdot (2i-1)^4 +1}$

The general form of all these expressions is $4 \cdot x^4 + 1$ . This can be factored out into $(2(x^2) - 2x +1) \times (2(x^2) + 2x +1)$ . Observe that $2 (x+1)^2 - 2 (x+1) + 1 = 2x^2 + 4x + 2 - 2x - 2 + 1 = 2x^2 + 2x + 1$ . Hence, we get that the product is equal to

$\begin{aligned} &\frac { \left[\prod_{n=1}^8 2(2n)^2 - 2(2n) + 1\right] \left[ \prod_{n=1}^8 2(2n+1)^2 - 2(2n+1) + 1\right]} { \left[\prod_{n=1}^8 2(2n-1)^2 - 2(2n-1) + 1\right] \left[\prod_{n=1}^8 2(2n-1)^2 + 2(2n-1) + 1\right]} \\ & = \frac { \prod_{k=2}^{17} 2(k)^2 - 2(k) + 1} { \prod_{n=1}^{16} 2(k)^2 - 2(k) + 1} \\ \end{aligned}$

As such, this product telescopes, and we obtain $\frac {2(17)^2 -2(17)+1}{2(1)^2-2(1)+1} = 545$ .

[Edits for clarity - Calvin]

Let's define $f(x) = x^4 + 64$ . We can factor f(x) like this: $f(x) = (x^2-4x+8)(x^2+4x+8)$ . Since all numbers in the expression taken to the fourth power are divisible by 4, we will make use of the function g(y) = f(4y). Then $g(y) = ((4y)^2 - 4(4y)+8)((4y)^2 + 4(4y)+8) =$ $= 64(2y^2-2y+1)(2y^2+2y+1)$ .

Let's define $h(y) = 2y^2-2y+1$ . Evaluating $h(y+1)$ , we get $2y^2+2y+1$ , which is the second nonconstant factor in the expression for g(y), so $g(y) = 64h(y)h(y+1)$ .

Thus, the expression we are asked to evaluate is $\frac{g(2)g(4)g(6)g(8)g(10)g(12)g(14)g(16)}{g(1)g(3)g(5)g(7)g(9)g(11)g(13)g(15)} =$ $= \frac{64^8h(2)h(3)h(4)h(5)h(6)h(7)h(8)h(9)h(10)h(11)h(12)h(13)h(14)h(15)h(16)h(17)}{64^8h(1)h(2)h(3)h(4)h(5)h(6)h(7)h(8)h(9)h(10)h(11)h(12)h(13)h(14)h(15)h(16)} =$ $\frac{h(17)}{h(1)} = 545$ .

I just calculated first 2 pairs of fractions and got 13/1, 41/13 respectively. Then, I noticed 1+(4+8)=13 and 13+(12+16)=41. Then I just kept writing rest of the pairs in this way and got 545.

(13/1).(41/13).(85/41).(145/85).(221/145).(313/221).(421/313).(545/421) = 545

We can use Texas Instruments Calculator as below.
- 6 store X
1 ENTER
$ANS~ store~ A : X + 8 ~store~ X : A*( (X+2)^4 + 64)/((X - 2)^4 + 64)$
Press ENTRY eight times we get 545
..(for ANS, ENTRY we have to use 2nd key, for A , and : the ALPHA key)

[(8^4+64)(16^4+64)(24^4+64)(32^4+64)(40^4+64)(48^4+64)(56^4+64)(64^4+64)] / [(4^4+64)(12^4+64)(20^4+64)(28^4+64)(36^4+64)(44^4+64)(52^4+64)(60^4+64)]