The Answer is Surprisingly Easy! | India National Mathematics Olympiad 2003

[yoav]

Notice that 1<a<2. Now we have by multiply the eq by a: a^6=a^4+2a-a^2
If we divide the eq by a: -a^2=2/a-1-a^4
So we get a^6=2a+2/a-1=f(a), with 1<a<2 . f'(a)=2-2/a^2=0 so a=1 f(1)=3 and f(2)=4 (you can check f(1/2)=4) so we see for1<a<2 we get 3< f(a)=a^6 <4.

[mathcanbeeasy]

And i came up with a simpler solution.
Ok, until 6/5<a<5/4 is classic. f continue, strictly increasing, f(6/5) is negative, f(5/4) is positive.
The next part is simpler and much beautiful.
If we multiply our equation with a we get a^6=a^4-a^2+2a
And we observe that if we multiply LHS again with a we can make something aproximate with original equation.
So, let's multiply again with a
a^7=a^5-a^3+2a^2
Now we add and subtract a and 2 in LHS to get the original equation.
a^7=a^5-a^3+a-2+2a^2-a+2.
So, a^7=2a^2-a+2
Now we divide by a to have again a^6
So, a^6=2a-1+2/a
Now is a part of simple algebra
6/5<a<5/4
12/5<2a<10/4
7/5<2a-1<6/4 (*)

6/5<a<5/4
4/5<1/a<5/6
8/5<2/a<10/6 (**)
Adding (*) and (**)
15/5<2a-1+2/a<19/6
So
15/5<a^6<19/6
So 3<a^6<3.17
And [a^6]=3

A little observation in final for the students which are not familiarised with olimpique exams:
Because f(1) is negative and f(2) is biiiiger way positive, is clearly that a is much closer to 1.
And is known by any olimpique student and is classic that the values for try are 3/2, 4/3, 5/4, 6/5, 7/6, 8/7 etc until f of one of this value is negative and f of the next value from the series is positive.
So this values(6/5 and 5/4) are not from nowhere, are from experience.

[carstenmeyer]

2:05: Multiply everything by *a^2 + 1* to simplify the polynomial and get an expression for *a^6* dependent on lower powers of *a*, so you can get away with the approximation *a ∈ (1; 2)* :

*a^7 + a = 2 * (a^2 + 1) => a^6 = 2 * (a + 1/a) - 1*

Note the RHS is increasing for *a > 1*, so we use *a ∈ (1; 2)* to approximate *a^6* via RHS

*a^6 ∈ (2 * 2 - 1; 2 * 2.5 - 1) = (3; 4) => ⌊ a^6 ⌋ = 3*

[federicoa]

Is a lot easier to notice that f(x)=-x^2+1+x-2=-x^2+x-1=0 mod(x^2-x+1) so f(x) is divisible by (x^2-x+1), do long division to find f(x)=(x^2-x+1)(x^3+x^2-x-2), and since the first factor is always positive, a has to solve a^3+a^2-a-2, and from that you can find by hand 3<a^6<4.

[Mephisto]

Audio and video are frequently desynchronized in your videos, you should look into it. Awesome content by the way!

[justinnitoi]

I just found a range of possible a values and continued to shrink the range until 3.06< a^6 < 3.138.

[holyshit]

Polynomial from this equation can be factored into
(a^2-a+1)(a^3+a^2-a-2)
This polynomial has only one real root
a^3=2+a-a^2
a^6 = (2+a-a^2)^2
a^6 = (a^2-(a+2))^2
a^6=(a^4-2a^2(a+2)+(a+2)^2)
a^6 = a^4-2a^3-3a^2+4a+4
a^6 = a^3(a-2)-3a^2+4a+4
a^6 = (2+a-a^2)(a-2)-3a^2+4a+4
a^6 =
a^6 = -a^3+4a
a^6 = -(2+a-a^2)+4a
a^6 = a^2+3a-2

[armacham]

A few things you can do. First use RRT to eliminate all rational roots, including all integers.

You showed any solution must be positive using derivatives, but you can also do it with factoring. split into three cases: x < -1, x is between (-1, 0), and x > 0. (knowing that there are no integer solutions we don't need to consider any other cases)

then you break the LHS into 3 parts and show that all 3 parts are negative, so they can't be equal to RHS which is zero

in the first case, where a < -1, you know that a < 0 and -2 < 0. And clearly a^5 - a^3 < 0

in the second case, where a is between (-1, 0), you know that a^5 < 0 and -2 < 0. And clearly a - a^3 < 0

so you can eliminate all negative solutions. knowing that a must be a positive irrational number makes things easier going forward

I also did the (a^2+1) thing to show that a^6 must be at least 3

to show that a^6 < 4 I did the following:

a(a^4 – a^2 + 1) = 2
a^4 - a^2 + 1 = 2/a
a^4 - 2a^2 + 1 = 2/a - a^2
(a^2-1)^2 = 2/a - a^2

this means 2/a - a^2 >= 0
2/a >= a^2
since we know a is positive we can multiply both sides by a without flipping the sign
2 >= a^3
4 >= a^6

thus a^6 must be between 3 and 4

[lt]

Great Video ! Like all the others. And sir, I would like to ask you a question : How do you make your videos ? On which software do you write your solutions? And audio ? Thank you in advance

[Эдвард-чэ]

I found lower bound like in the video, and the upper bound this way: a³+2=a⁵+a⩾2a³ => a³⩽2 => a⁶⩽4, but equality holds only when a⁵=a, so a=0, ±1 which are not roots. So a⁶<4.

[Utesfan]

At the 4 minute mark you had bound 5/4<a<5/6, hence 2<a^6<4.
(likely 3/2 and 4/3 were tried off camera first)

Let x=3^(1/6). We need to determine the sign of f(x).

x^5 - x^3 + x-2 <?> 0, rearrange, using x>0:
x^4 -x^2 +1 <?> 2/x multiplying by x^2+1:
x^6-1 <?> 2*(x^2+1)/x using the definition of x:
2 <?> 2(x^2+1)/x.

But (x^2+1)/x = x + 1/x > x for all positive real numbers, including 3^(1/6).
Hence f(x)<0, and x<a<5/6 and 3<a^6<4.

Thus Floor(a^6)=3.

[bait]

Is it safe /robust just to find a in (1, 2)...then do the alg-manip to bound a^6 with the lower bound
of a/(a^2+1) at a=2 thus bounding 3<=a^6<4.
Thus skipping the fractional bounding?

[philippenachtergal]

0:00 Hum.
f(0) = -2
f(1) = -1
f(2) = 24
So there is a root between 1 and 2 which isn't precise enough.
now, f'(x) = 5a^4 - 3a^2 + 1, f'(1) = 3
So first iteration of newton's method tells me to try for a root around 1 - (-1)/3 = 1.3333
If we test at 1.3 we get f(1.3) > 0
If we test at 1.2 we get f(1.2) = -0.03968
1, 2 ^ 6 = 2, 985984 so argh we are so close to 3 that I can really say if floor(a^6) is 2 or 3
Not without doing calculs that would be quite annoying to do by hand
Now, if I were to pretend that I would do this kind of calcul by hand, I would eventually figure out that f(1.201) < 0 and f(1.21)>0 so there is a root between those two.
As 1.201^6 > 3 and 1.21^6 < 4 we have floor(a^6) = 3
Answer is 3
(at least one of the answers that is, there might be other roots)

But yeah, I wouldn't do that by hand unless I was paid handsomely for it.
So another approach is needed.

[mojota]

The trick at 5:18 seems to come out of thin air. Can you give some insight into how you came up with it?

[bait]

Some calculus and numerical bracketting nice

[wonjonghyeon]

At least for me, it is not an easy solution...

[petersievert]

If you invest a lot of time and sweat, you could come up with 241/200 < a by calculating the according values and accordingly 3 < (241/200)^6 < a^6 .
But quite in the contrast I'd rather be interested if it is possible to show that f( 6th root of 3) < 0 and thus a > 6th root of 3 .
It was rather easy to show f (6th root of 4) > 0, so a solution like that would seem a reasonable approach to me.

[Ritesh_.]

Ive done using calc, i ruined the beauty of the q😢

[predator]

Please explain what you are writing....!!!

[deejayaech]

Use newtons method to find the real root of the polynomial to a good degree of approximation and from there floor(a) is trivial