_x = ½(1 + √5) = ½( -(-1) + √( (-1)² - 4(1)(-1) )_ ∴ _x² + (-1)x + (-1) = 0_ ⇒ _x² = x + 1_ Multiply through by _x:_ _x³ = x² + x = (x + 1) + x = 2x + 1_ ⇒ _x⁶ = (2x + 1)² = 4x² + 4x + 1 = 4(x + 1) + 4x + 1 = 8x + 5_ ⇒ _x¹² = (8x + 5)² = 64x² + 80x + 25 = 64(x + 1) + 80x + 25 _ _= 144x + 89 = 72(1 + √5) + 89_ ∴ *_x¹² = 161 + 72√5_*

Another proof : noticing that (1+ √5)/2 = φ, and φ^2 = 1 + φ, we have the property φ^n = Un-2 + φ Un-1 for all n ≥ 2, where (Un) is the Fibonacci sequence. Then φ^12 = 89 + 144 φ = 161 + 72 √5. If you do not like finding the Fibonacci elements until n = 11, just take φ^6 = 5 + 8 φ, and square φ^6 : φ^12 = (φ^6)^2 = (5 + 8 φ)^2 = 89 + 144 φ = 161 + 72 √5. Thank you for your videos !

This solution is far too cumbersome. (((Term^2)^2)^3 is much faster

I think it’s simpler just to expand out as ((((1+sqrt 5)/2)^2)^2)^3 as the expression squares out pretty simply each time

I was screaming, 'GET ON WITH IT!', Lol.

There’s another elegant way to solve this problem by recognizing that (1+sqrt(5))/2 is the eigenvalue of the matrix A = [0,1; 1, 1] with largest magnitude. Taking this matrix to powers produces matrices with Fibonacci numbers as entries A^p = [f(p-1), f(p); f(p), f(p+1)], so A^12 = [89, 144; 144, 233]. The largest eigenvalue of A^12 is ((1+sqrt(5))/2)^12. Simplifying the characteristic equation for A^12 gives 0 = x^2 - 322*x + 1 and computing the larger root gives the answer.

φ^z = F(z-1) + F(z)*φ holds true for any integer power of φ, negative, zero or positive. F(z) is the Fibonacci numbers with index z. To find the Fibonacci numbers with negative index you can reverse the formula to F(z) = F(z+2) - F(z+1) to get the sequence F(-1)=1, F(-2)=-1, F(-3)=2, F(-4)=-3, F(-5)=5, F(-6)=-8 and so on, i.e the same as the positve indexed Fibonanacci numbers, but negative if the index is even. Plug in 12 for z and you will get: φ^12 = F(12-1) + F(12)*φ = F(12-1) + F(12)*φ = F(11) + F(12)*φ = 89 + 144*φ which can be simplified to F(z-1)+F(z)/2 + (F(z)/2)*sqrt(5) if F(z) is even.

Such a great opportunity missed, to note in passing that [ (1 + √5) / 2 ]³ = (1 + 3√5 + 3*5 + 5√5) / 8 = (16 + 8√5) / 8 = 2 + √5; and thus give students insight into this instance (and others similar) of the Pell equation. Now the entire expression simplifies as [ (1 + √5) / 2 ]¹² = (2 + √5)⁴ = (4 + 4√5 + 5)² = 81 + 72√5 + 80 = 161 + 72√5

It's the golden ratio. That said, knowing the golden ratio and the afferent properties (x=1+1/x) is more a matter of general math culture, and olympiads are for teenagers. some of them will know and some won't at all, this is NOT the spirit of the olympiads. Those problems should be based on problem-solving ability, not on random general culture.

We only have to calculate φ^2, φ^4, φ^8 by consecutive squaring and then φ^12 =φ^4×φ^8. It's just four lines.

It is not clear to me what is simpler, the original equation or the calculated result.

i don't think this video should be 11 minutes long

You complexified everything my friend. My idea was that we could detect a nice generalisation for Sn=x^n that would not force us to write Newton's formula with a coeff of 12. So I started looking at S1=(1+sqrt(5))/2, S2=(3+sqrt(5))/2, S3=2+sqrt(5). And now my idea of a nice formula vanished but S3 is so simple that it can be used to compute easily S6=S3²=9+4sqrt(5) and then S12=S6²=161+72sqrt(5)

Want to have a shortcut? Fib(12) = 144. Considering the explicit form of Fib sequence, φ^12=144*sqrt(5) + ((1-sqrt(5))/2)^12 which is approximately equal to 144*sqrt(5) or 322.

φ = (1+√5)/2 where φ is the golden ratio φⁿ=Fₙ₋₁+Fₙφ where (Fn) denotes the Fibonacci sequence φ¹²=F₁₁+F₁₂φ with F₁₁=89 and F₁₂=144 φ¹² =89+144((1+√5)/2)=161+72√5

Here's what everyone is missing: the solution demonstrated would be doable by anyone who has done basic algebra

z = ½(1 + √5) is a root of x²-x-1, so z²=z+1. Now it is easy to calculate z^12 stepwise and while doing so replace each occurence of z² by z+1. This gives z^12=144z+89. Substituting z = ½(1 + √5) gives 161 + 72√5.

A good teacher doesn’t just show a series of steps. A good teacher explains the goal and explains why each step works toward the goal. Otherwise this is just rote learning and not knowledge.

The point of the question is to determine whether students are aware of the significance of reducing the power demanding evaluation. This is allows programming with minimal error growth. (In this instance the Harmonic ratio and Fibonacci sequence are useful, and may be part of a bonus, but the most significant factor is spotting ways to minimize error growth in numerical calculations. )

Without watching: (1+5^½)/2 is the constant phi (sorry, no Greek keyboard here). Phi ^ n = Fn×phi + Fn-1, where Fn is the nth number in the Fibonacci series. Moreover, this is approximately equal to Fn+1 + Fn-1. So, for power 12, this would be 233+89=322. The actual number is more like 319.99.

Surely the 'answer' here is simply a different way of writing the question?

½(1 + √5) = x x^12 = (x^2)^6 x^2 = t x^12 = (t^2)^3 and just calculate

When you know some advanced geometry with the golden ratio, it is simple. (1+sqrt(5))/2 = phi, so phi^2=phi+1 square it: phi^4=phi^2+2phi+1=3phi+2 square again: phi^8=21phi+13 multiply last two results together: phi^12=phi^8*phi^4=(21phi+13)(3phi+2)=144phi+89=72*sqrt(5)+161. If you don't know the basic property of the golden ratio, then one will not get the first idea in this video. If you do, you can immediately start with step 2: x^2=x+1

A beautiful solution. I did it without x at first, carefully, and got it right.After that I could finally understand your simpler approach , and I then did it that way too. Thank you.

Two Fibonacci series, each with different f(1) and f(2) values. 1,3 and 1,1. The latter can also be derived from (1/sqrt(5))*[((1+sqrt(5)/2)^n - ((1-sqrt(5)/2)^n]

By factoring 12 to 3*2*2, raising to the power of 12 can be done by cubing, followed by squaring twice. Let x = √5 + 1 We have x³ = 5*√5 + 3*5 + 3*√5 + 1 = 8*√5 + 16 Thus, ((√5 + 1)/2)¹² = (((x/2)³)²)² = ((x³/8)²)² = (((8*√5 + 16)/8)²)² = ((√5 + 2)²)² = (4*√5 + 9)² = 72*√5 + 161

phi = x=(1+sqrt(5))/2; phi^n=F(n)phi+F(n-1), where F(n) - the n-th Fibonacci number: 1,1,2,3,5,8,13,21,..... F(n)=F(n-1)+F(n-2). The same another way: x^0 = 1 = 1+0*sqrt(5) = (1;0) x^1 =(0.5;0.5) x^n=(a(n);b(n)) = (a(n-1);b(n-1))+ (a(n-2);b(n-2)).

Hard because the question is incomplete. Some sort of result format specification needed. Decimals? Rational number format etc

(1-√5)/2との和と積から解と係数の関係の逆からx^2-x-1=0は出せる。

Needs calculations but is straightforward: ((1+ √5)/2)^12 = ((1+√5)x1/2))^12 = ((1+√5)x1/2))^2x6 = [(1+√5)^2)x (1/4)]^6. then we apply the (a+b)^2 formula = a^2 + 2ab + b^2... given that (a+b)^6 = (a+b)^(2x3) = ((a+b)^2)^3... we come to a point where we have... [1+√5(2+√5)]^3 x 1/(4^6).... we apply the rule (a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3... the rest is pure calculations... Many calculations but no tricks!!!! straightforward answer. Finaly we multiply with 1/(4^6), basically division.

It's much better to firstly calculate x^3=2+√5 because it immediately gets rid of the /2. Then find the solution from (2+√5)^4 via two successive squarings.

Alternative? extract even and odd powers of expansion (1+x)^12 = p1(y) +x*p2(y) here x=sqrt(5), and y=x^2=5 then divide by 2^12. in this case p1(y)=y^6 + 66*y^5 + 495*y^4 + 924*y^3 + 495*y^2 + 66*y + 1 = 659456 -> 161. p2(y)=12*y^5 + 220*y^4 + 792*y^3 + 792*y^2 + 220*y + 12 =294912 -> 72=> (p1(y)+x*p2y)/2^12 = 161+72*sqrt(5)

Small deduction for ambiguous "1" vs. "7" display. Yet a miracle occurs, that "c" and its inversion somehow produce "x", and it's clear and unambiguous. Every time I saw that I both winced and smiled.

I bet the actual exam had only 5 lines to do working in lmao

A very elegant solution. (That is, if the year is before 1970 and no scientific calculators are available.) 161 + 72 * sqrt( 5 ) does indeed equal ( ( 1 + sqrt( 5 ) ) / 2 ) ^ 12

Moivre-Binet formula for implicit Fibonacci solves :)

0:00 if you solve for the terms within the parentheses you get the golden ratio 🤩

Trying to keep it simple (to be safe) I just cubed the expression then squared it twice. Cubing gets rid of the denominator so it works out pretty nicely.

Prop. Phi = x , x ^n = F(n) x + F(n-1) = F(n)/2 + F(n-1) + + F(n)/2 * 5^.5 F(1) = F(2) = 1 F(n) = F(n-1) + F(n-2) F(12) = 144, F(11) = 89

(1 + √5)/2 is the golden ratio, and it has some properties related to the fibonacci number. I will call this number g(for golden ration). First of all, we have g²=g+1; therefore, if we multiply g both sides, we get g³ = g² + g, and on we go. Now if we replace g², we end up with g³ = g + 1 + g, that being g³ = 2g + 1. If we do it recursively, you will notice that this expression will go like: gⁿ = Fibo(n)g + Fibo(n-1) being Fibo(k) the k-th number of the fibonacci sequence. For those who never saw this before, the fibonacci sequence is a sequence where the next term is the sum of the last 2 terms. It starts at 1. To get to the next term, we sum 1+0, because we have nothing before the starting term, then we get a 1. For the 3rd term, we sum 1+1 and get 2, being 1 and 1 the last 2 terms. For the next, we sum 1+2 and get 3, being 1 and 2 the last 2 terms, and so on. The Fibonacci sequence should look like this: Fibonacci {1, 1, 2, 3, 5, 8, 13, 21, 34, 55}. What is so important about this sequence is that if we get the ratio of 2 consecutive terms, we approach the golden ratio(named g before). now if we want to get the value of g¹², we need to get the value of Fibo(12) and Fibo(11), those values being 144 and 89. We end up with g¹² = 144g + 89. replacing (1 + √5)/2 in g, we get 161 + 72g

the way he wrote 1 is weird

8% of students got this right. 98% of the rest of humanity thought, "Why do we pay these people for this pointless gibberish?"

Use the binomial theorem

X is two straight lines crossing. What is this atrocity you are drawing? :)

You overcomplicated this poor little thing. Once you wrote that x²=x+1 you have: x^3=x².x=(x+1).x=x²+x=2x+1 x^6=(2x+1)²=4x²+4x+1=4.(x+1)+4x+1=8x+5 x^12=(8x+5)²=64x²+80x+25=64.(x+1)+80x+25=144x+89 So the value we are looking for is 72.(1+sqrt(5))+89=161+72.sqrt(5)

x²-x-1=0 x²-x+1=2 x³+1=2x+2 x³=2x+1=2+sqrt5 x⁶=9+4sqrt5 x¹²=161+72sqrt5 수능 15번보다 쉬운듯

Thank you for showing me how to reach the result with all the substitutions.

Thanks for helping learn math. discover the math of reality. I stopped because I already know the answer.

Smart guys unattempt question and saves time😎

I don't know if I'm more impressed by the maths or the neat handwriting.

If you consider that (1+(5)^.5)/2=Phi, the golden ratio, and the relation Phi^n=Phi[n]*Phi+Phi[n-1], where Phi[n] is the n-th number of the Fibonaci sequence. Thus, Phi^12=Phi[12]*Phi+Phi[11], and finally, Phi^12=144*Phi+89.

If you happen to recognize ϕ, and know this one little trick, along with the first dozen Fibonacci numbers, you can do this one in seconds in your head: ϕⁿ = F(n)ϕ + F(n-1) In the given problem, n = 12, so we can write (½[1 + √5])¹² = ϕ¹² = F(12)ϕ + F(11) = 144ϕ + 89 = 72[1 + √5] + 89 = 161 + 72√5 And if you don't know the first dozen Fibonacci numbers, you can quickly produce them by the Fibonacci recursion rule, along with F(0) = 0, F(1) = 1: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 Fred

((1+√5)/2)^12 = phi^12 we know phi^2 = 1 + phi phi^12 = phi^(2×6) = (phi^2)^6 = (1+phi)^6 = ((1+phi)^2)^3 = (1+2phi + phi^2)^3 = (1+2phi+1+phi)^3 = (2+3phi)^3 = (2+3phi)^2 (2+3phi) = (4+12phi+9phi^2) (2+3phi) = (4+12phi+9(1+phi)) (2+3phi) = (4+12phi+9+9phi)(2+3phi) = (13+21phi) (2+3phi) = 26 + 39phi + 42phi + 63phi^2 = 26 + 81phi + 63(1+phi) = 26 + 81phi + 63 + 63phi = 89 + 144phi = 89 + 144((1+√5)/2) = 89 + (144/2) (1+√5) = 89 + 72(1+√5) = 89 + 72 + 72√5 = 161 + 72√5 so, ((1+√5)/2)^12 = 161 + 72√5

Modern students will not understand why the final solution is any simpler than the initial problem. This is understandable, because the point of solving problems without a calculator is debatable. If only 8% are going to get the right answer, then why do it? They are more likely to have a calculator than a book of Mathematical Tables or a slide rule, the use of which was the object of these exercises when I was at school fifty years ago.

Awesome, thanks. Very helpful.

Isn't it simple ✓5 = 2.2360 then +1 and whole ÷ 2 and then 1.618 to the power 12 = approx 322

easy for anyone knowed or can see that x²=x+1.. Many examples in calculus are tributary to some anterior knowledge. Utility write any number as solution of some equation than try to simplify it by this equation Particular case :if there is sq root it will be root of second equation Ex: x=1-√5 (1-√5)¹⁰? (1-√5)²=1²-2*√5+5=6-2√5=4+2(1-√5) =4+2x x²=4+2x x⁴=16+16x+4x²=16+16x+4(4+2x) =32+24x x⁸=(32+24x)² =32²+2*32*24x+576 =1600+1536x x¹⁰=(1600+1536x)*(4+2x) =6400+3200x+6144+3072x =12544+6272x =12544+6272*(1-√5) =6272(2+1-√5)) =6272(1-√5) Finally x¹⁰=6272x

We know for a = phi: a²=a+1 so a³=a²+a=2a+1 = 2+ V5 a^12 = (2+ V5)^4 = (9+4V5)² = 161 + 72V5

A more elegant approach would be to iterate. After noting the expression to be raised to the power of 12 is a root of x^2 - x - 1, it follows that x^2 = x + 1. So x^3 = x^2 + x = 2x + 1. And therefore by iteration any power of x can be expressed in terms of ax + b. If x^n = ax + b then x^(n+1) = (a+b)x + a. You can do the rest in your head.

Bro I think that You have to use binomial expression to solve this question. In India this math problem are simple . In India High school student slove this simple problem.

The worst part is: your pen is not working !

Just know that phi²=phi+1 and boom, ez solution, just develop (phi+1)², replacing phi² by phi+1, and then do it with (a*phi+b)³ and you've got it

After you arrived at (3x + 2)^3 , It would be more obvious to just do the cubic binomial expansion (a+b)^3= a^3 + b^3 + 3a^2b + 3ab^2; also the Tartaglia triangle was a viable method to do that binomial expansion!

F = PHI F = ( sqrx( 5 ) + 1 ) ÷ 2 F^2 = F + 1 F^6 = ( F^2 )^3 = ( F + 1 )^3 .......... F^6 = 8F + 5 F^12 = ( F^6 )^2 = ( 8F + 5 )^2 ............. F^12 = 144F + 89 = 72 sqrx( 5 ) + 161 👍😁🤪👋

It's perhaps easier to figure out first the relation phi^n = F_n . phi + F_{n-1} where F_n is the n-th Fibonacci number; then compute phi^12 directly.

Only 8% of 6th graders I assume?

Sir, I would like to take your views on solving (24x+13)(3x+2) Here, 24x*3= 72x By the way mind blowing lecture❤

I just did this by repeated application of x^2, x^4, x^8 ; then x^4 * x^8, using x^2 = 1+x φ^2 = 1 + φ φ^4 = (φ^2)^2 = (1 + φ)^2 = 1 + 2φ + φ^2 = 1 + 2φ + (1 + φ) = 3φ + 2 φ^8 = (φ^4)^2 = (3φ + 2)^2 = 9φ^2 + 12φ + 4 = 9(1+φ) + 12φ + 4 = 21φ+13 φ^12 = (21φ + 13)(3φ+2) = 89 + 144φ

Ce nombre est φ, le nombre d'or et on sait que φ est solution de l'équation x² - x - 1 = 0 donc: φ² - φ - 1 = 0 d'où: φ² = φ + 1 φ¹² = (φ²)⁶ = (φ + 1)⁶ = [ (φ + 1)²]³ = ( φ² + 2φ + 1)³ = (φ + 1 + 2φ + 1)³ = (3φ + 2)³ = (3φ + 2)² (3φ + 2) = (9φ² + 12 φ + 4) (3φ + 2) = (9φ + 9 + 12 φ + 4) (3φ + 2) = (21φ +13) (3φ + 2) = 63φ² + 42φ +39φ +26 = 63φ + 63 + 42φ +39φ +26 = 144φ + 89 = 144 (1 + √5)/2 + 89 = 72 (1 + √5) + 89 = 72 + 72√5 + 89 = 161 + 72√5

Why the equal numbers to X. Isn't it 20^6

3 sol. with x=(1+r5)/2 Common x2=(1+2r5+5)/4=(6+2r5)/4=(3+r5)/2 x4=x2*x2=(9+6r5+5)/4=(14+6r5)/4=(7+3r5)/2 x8=(49+42r5+45)/4=(94+42r5)/4=(47+21r5)/2 x12=x8*x4=(329+141r5+147r5+315)/4=(644+288r5)/4=161+72r5 Better x2=(1+2r5+5)/4=(6+2r5)/4=(3+r5)/2 x4=x2*x2=(9+6r5+5)/4=(14+6r5)/4=(7+3r5)/2 x6=x4*x2=(21+7r5+9r5+15)/4=(36+16r5)/4=9+4r5 x12=x6*x6=81+72r5+80=161+72r5 Easier x2=(1+2r5+5)/4=(6+2r5)/4=(3+r5)/2 x3=x2*x=(3+3r5+r5+5)/4=(8+4r5)/4=2+r5 x6=x3*x3=4+4r5+5=9+4r5 x12=x6*x6=81+72r5+80=161+72r5

I was always told that the number was *tau* - the Golden ratio - not phi . Often encountered in nature. But either Greek letter will do, I suppose.

I immediately set the golden ratio to x and jumped to x^2 - x - 1 = 0 => x^2 = x + 1. Multiply both sides by x, simplify, square, simplify, square again, simplify, substitute, and solve, abusing x^2 = x + 1 whenever possible.

No need to use x. Just split the numerator and denominator, then square the numerator three times getting the common factor out every time simplifying with the denominator, and finally multiply one more time the numerator by 7+3sqrt(5) to get 644/4 and (288/4)sqr(5) or 161 + 72sqr(5). That simple.

I solved by factoring out 2 in step 1, this gives 2^-12 as a factor and (1+sqrt(5)) power 12 Now in each step you will get more factors that you can take out and power. Its almost the same amount of work that way.

You break down the 12th power into 2*2*3, taking x, squaring it twice, and then taking the cube. I would have taken the cube up front; x^3 = x^2*x = (x+1)*x = x^2+x = 2x+1, and then squared that twice instead. Or, observing that your x is the golden ratio φ, I would recognize that φ^n = (L(n) + F(n)*sqrt(5))/2 where L(n) and F(n) are the Lucas and Fibonacci series respectively. One can work out the series manually from the recursive definition, which for n=12 is feasible to do directly. For larger n, the methods for skipping ahead in these series are fundamentally the same as the manipulations shown in this video (multiplying two exponents).

(1+sqrt5)^12=(((1+sqrt5)^3)^2)^2 =((1+3sqrt5+15+5sqrt5)^2)^2 = ((16+8sqrt5)^2)^2= ((8(2+sqrt5))^2)^2 = 8^4 x(4+4sqrt5+5)^2 = (2^3)^4 x(9+4sqrt5)^2 = 2^12 x(81+72sqrt5+80) = 2^12 x(161+72sqrt5), and 2^12/2^12 = 1, and the result is 161+72sqrt5

Square it. Square that (^4). Do it one more time (^8) Multiply the 4th power and 8th power results. A little simplification and you get to the answer in about 1/4 the time.

((1+sqrt(5))/2)^3 = 2+sqrt(5) (2+sqrt(5))^2=9+4*sqrt(5) (9+4*sqrt(5))^2=161+72*sqrt(5) This took me 2mins to compute.

this expression is φ (The Golden ratio) raised to the 12 power. Then, φ² = 1 + φ.

I love how every overinflated ego posts a different solution using more and more advanced mathematics, criticizing his approach, but no one cared to ask who this video was made for. You don't teach how to solve such a problem the same way when your target audience is high school students as you would if it were made for people pursuing a university degree in mathematics. Some of his videos have "Entrance exam" in the title, for different universities, which implies that this may be for people who don't have all the techniques some of you are talking about.

Once you have established x² = x + 1 with x≠ 0, just multiply by x to get an equation in terms of x³: x³ = x² + x = (x + 1) + x = 2x + 1 From here keep going to get equations for each integer exponent: x⁴ = x³ + x² = (2x + 1) + (x + 1) = 3x + 2 x⁵ = x⁴ + x³ = (3x + 2) + (2x + 1) = 5x + 3 x⁶ = x⁵ + x⁴ = (5x + 3) + (3x + 2) = 8x + 5 x⁷ = x⁶ + x⁵ = (8x + 5) + (5x + 3) = 13x + 8 x⁸ = x⁷ + x⁶ = (13x + 8) + (8x + 5) = 21x + 13 etc. The reason for the Fibonacci sequence coming into play that other posters have pointed out becomes apparent.

It took me 2 minutes after rewriting (1+√5)/2 in four groups to the power of 3. By using once the formula of the cube, once the formula of the square and then rearranging the resulting number, you get the result.

Actually it's hard but I brove it by binomial theorem ☝️🤓

A rule for solving this sort of problem: If it's too much like hard work, you've missed a trick... It's easy to see that... x^2 = x+1, so x^4 = 3x+2, so x^6 = 8x+5, so, finally x^12 = 144x+89 Then the result is easy to compute.

This questions is lot easier if you recognize this is φ

I almost went into a rant about how arithmetic over Z[phi] could be hard, but... you know, when 8% can solve it, it's a very simple problem, because 95% usually didn't even try.

Square by hand to obtain φ squared Multiply again by φ to obtain 3rd power Square the 2nd power to obtain 4 power Cube the 4th power

You could get (3x + 1)^3 via the binomial expansion. I.e. Pascal's triangle.

(½(1 + √5))¹² = (¼(1 + 2√5 + 5))⁶ = (¼(6 + 2√5))⁶ = (½(3 + √5))⁶ = (¼(9 + 6√5 + 5))³ = (½(7 + 3√5))³ = ⅛(343 + 441√5 + 945 + 125√5) = 161 + 72√5 I can't get it why you made it so complicated?

I thought your approach required minimal Algebra knowledge, so I thought it was an excellent approach. Some of the other solutions below were shorter but required more math knowledge to understand.

Interesting question. simple answer. thx for the vid

I'm not a mathematician. But after your many manipulations, what is the actual result? Because your final line is not much simple from the original line. If you do not simplified, what was the point to do all these manipulations? You can just solve the original one on a calculator.

I was a software engineer for forty years. If this guy was working for me I would have fire him. I know this "guy" - he is far more interested in proving how smart is than actually getting any work done.

He took so long I forgot the original problem

Decent presentation of the basics, but its a great opportunity to demonstate tge interconnectedness of ideas here. Once you have tgat expression fir x^2, yiu van get one from x^3. Then by breaking down non kinear terms into linear exoressions, you can save time. So one cannedplicitky introduce an iterative method. Isnt this closely tier to the fibonacci series?

So from memory from 30 yrs ago when I studied mathematics, isn’t this about a group modulo an ideal or something?

What cultures write x with two curves rather that lines that cross?

Since x²=x+1, then x⁴=(x²)² =(x+1)²=x²+2x+1 =x+1+2x+1=3x+2 x⁸=(x⁴)² =(3x+2)²=9x²+12x+4 =9(x+1)+12x+4=21x+13 x¹²=(x⁴)(x⁸) =(3x+2)(21x+13) =63x²+81x+26=63(x+1)+81x+26 =144x+89 Since x=(1+√5)/2, then x¹²=144(1+√5)/2+89 72+72√5+89=161+72√5

The comments are way more informative and interesting than the video.