To establish your bounds you're implicitly assuming your b and c have the same sign. This is no problem after possibly reordering the variables but should be said
@@letsthinkcritically Subject to your argument for 4:22 I wanted to point out that b,c being roots to x²+ax+(a²-2011) can't imply that b,c are of equal sign since we would need to show that a²-2011>0 which is true but is shown later. I believe one should start how Austin suggested (pigeon principle on the signs of a,b,c and re-order accordingly).
Thank you for pointing this out. I pause the video at this exact spot, thinking “you can’t assume that’s always true, can you?“. I had to rewatch that part of the video several times to make sure I didn’t miss something that was said, but not written, to justify the any quality that was “clearly “true. But, as added in the description, of the three integral roots a, B, C, at least two of them must have the same signs, and hence you may assume without loss of generality that b and c have the same signs.
Differentiate to find a local minima and maxima. Where x= +/- sqrt(2011/3). There must be a root between - 25 and 25. Thus the perfect square has to be between 8044 and 8044-3(625). Test cases between 78 and 89 for sqrt of the perfect sq. Note 8044 is 1mod3, (you could have used this as well) thus the perfect square must be 1mod3. Disregard multiples of 3 for the perfect square.
Everything from 4:00 to 7:00 would go more smoothly if you first rewrote the quantity in terms of b+c and b-c. b^2 + bc + c^2 = 3/4*(b+c)^2 + 1/4*(b-c)^2
From a problem construction view, is it usual to ask for something like |a|+|b|+|c|, when the way of doing this is to find a, b, and c! I was hoping for some clever approach that gave a way of deriving the sum of absolutes without finding the roots!
If a,b,c are roots of polynomial then c=-(a+b). By Vieta's identities -2011=a*b+a*c+b*c=a*b+(a+b)*c=a*b+(a+b)*(-a-b)=-a^2-a*b-b^2. Then a^2+a*b+b^2=2011. 2011=(a-b*w_1)*(a-b*w_2) where w_1, w_2 are roots of x^2+x+1. But 2011 is a rational prime. Then (a-b*w_1) is a prime in Z[w_1]. But 2011=44^2+3*25=44^2+3*5^2=(44+5*√(-3))*(44-5*√(-3)). Then we can use that (a-b*w_1)=(44+-5*√(-3))*(-w_1)^k (**) where k=0,1,2,3,4,5. Because unities in Z[w_1] are the form (-w_1)^k where k=0,1,2,3,4,5. Indeed we can suppose the √-3=2*w+1. Now solving (**) is easy.
Since this polinom is an increasing-decreasing-increasing one, with max at x=-25.9 app and min at x=25.9 app, I wrote a c program and found -49 , 10 and 39 to be also a solution. Final solution is the same.
From Vieta's formulas we have abc = -m for this problem. So |m| = |a|*|b|*|c| = 49*39*10 = 19110 There are 4 solutions for (a,b,c): (49,-39,-10),(49,-10,-39),(-49,39,10),(-49,10,39). m=+-19110 depending on which solution you take. Quite astonishing that there are only 2 choices for integer m that will allow the cubic to have integral roots. The integral root condition is extremely restrictive!
There can be 2 positive roots and one negative. Please explain what you're doing. Even in the description your arguments aren't understable (as well as false, so...). Say "at least two of the roots have the same sign. Without loss of generality, let's assume that this is the case of b and c". Same for a>0, you can assume that but you have to say that if a triple (a,b,c) is a solution then so is (-a,-b,-c). And when you're looking for lambda, how are we supposed to understand your thought? You just wrote and said that lambda satisfies the inequality. Say that you're going to prove such an inequality and proceed by equivalence while writing and saying "" or "iff" at each step. Lastly, AM-GM is really overkill to prove a^2+b^2 ≥ 2ab, developping (a-b)^2 is simple and straightforward
the lower bond can be directly written if one knows about the Sophie inequality it states that
3/4(a+b)² ≤ a² + ab + b²
To establish your bounds you're implicitly assuming your b and c have the same sign. This is no problem after possibly reordering the variables but should be said
Please refer to the remark in the description.
@@letsthinkcritically Subject to your argument for 4:22 I wanted to point out that b,c being roots to x²+ax+(a²-2011) can't imply that b,c are of equal sign since we would need to show that a²-2011>0 which is true but is shown later.
I believe one should start how Austin suggested (pigeon principle on the signs of a,b,c and re-order accordingly).
At 4:22, when is the assumption made that both b and c >= 0? The inequality does not necessarily hold.
Sorry that I skipped the argument. Please refer to the remark in the description.
@@letsthinkcritically Thanks. Love the content!
You r are right !!!!
We need magnifying glass to read your writing!
Thank you for pointing this out. I pause the video at this exact spot, thinking “you can’t assume that’s always true, can you?“. I had to rewatch that part of the video several times to make sure I didn’t miss something that was said, but not written, to justify the any quality that was “clearly “true. But, as added in the description, of the three integral roots a, B, C, at least two of them must have the same signs, and hence you may assume without loss of generality that b and c have the same signs.
Differentiate to find a local minima and maxima. Where x= +/- sqrt(2011/3). There must be a root between - 25 and 25. Thus the perfect square has to be between 8044 and 8044-3(625). Test cases between 78 and 89 for sqrt of the perfect sq. Note 8044 is 1mod3, (you could have used this as well) thus the perfect square must be 1mod3. Disregard multiples of 3 for the perfect square.
The move of setting a boundary for a is brilliant! I started from 1 to test 8044-3b2 is a perfect square.
Luckily the first answer is 10😂😂
You can make video on the functional equation question asked in inmo yesterday it was very interesting
Everything from 4:00 to 7:00 would go more smoothly if you first rewrote the quantity in terms of b+c and b-c.
b^2 + bc + c^2 = 3/4*(b+c)^2 + 1/4*(b-c)^2
From a problem construction view, is it usual to ask for something like |a|+|b|+|c|, when the way of doing this is to find a, b, and c! I was hoping for some clever approach that gave a way of deriving the sum of absolutes without finding the roots!
I think the point is that if a, b & c are solutions then so are their opposites, so the sum of absolute values removes the ambiguity.
If a,b,c are roots of polynomial then c=-(a+b). By Vieta's identities -2011=a*b+a*c+b*c=a*b+(a+b)*c=a*b+(a+b)*(-a-b)=-a^2-a*b-b^2. Then
a^2+a*b+b^2=2011. 2011=(a-b*w_1)*(a-b*w_2) where w_1, w_2 are roots of x^2+x+1.
But 2011 is a rational prime. Then (a-b*w_1) is a prime in Z[w_1].
But 2011=44^2+3*25=44^2+3*5^2=(44+5*√(-3))*(44-5*√(-3)).
Then we can use that
(a-b*w_1)=(44+-5*√(-3))*(-w_1)^k (**) where k=0,1,2,3,4,5. Because unities in Z[w_1] are the form (-w_1)^k where k=0,1,2,3,4,5.
Indeed we can suppose the √-3=2*w+1. Now solving (**) is easy.
Whybuse Vieta at all and not just write mx plus a times nx plus b times wx plus c and work out like that??
OP !
Its really confusing you use a,b,c for the general equation en donde b is 2011 you use a,b,c in the symmetric sum why do you place p q,r apart?
Since this polinom is an increasing-decreasing-increasing one, with max at x=-25.9 app and min at x=25.9 app, I wrote a c program and found -49 , 10 and 39 to be also a solution. Final solution is the same.
Can you share your Program please.
I see that the values of a,b, and c are found but... what is the value of m?
From Vieta's formulas we have abc = -m for this problem.
So |m| = |a|*|b|*|c| = 49*39*10 = 19110
There are 4 solutions for (a,b,c): (49,-39,-10),(49,-10,-39),(-49,39,10),(-49,10,39).
m=+-19110 depending on which solution you take.
Quite astonishing that there are only 2 choices for integer m that will allow the cubic to have integral roots.
The integral root condition is extremely restrictive!
m=2011x-x^3 just draw graph..
U took a as negetive wlog ...but this concludes that b and c are also negetive then how they added up to give 0
I don't understand why b2+bc+c2 less than (b+c)2? How do you know bc is great than 0?
B and C are negative (check the description)
I'm first , I have done this problem
There can be 2 positive roots and one negative. Please explain what you're doing. Even in the description your arguments aren't understable (as well as false, so...). Say "at least two of the roots have the same sign. Without loss of generality, let's assume that this is the case of b and c". Same for a>0, you can assume that but you have to say that if a triple (a,b,c) is a solution then so is (-a,-b,-c). And when you're looking for lambda, how are we supposed to understand your thought? You just wrote and said that lambda satisfies the inequality. Say that you're going to prove such an inequality and proceed by equivalence while writing and saying "" or "iff" at each step. Lastly, AM-GM is really overkill to prove a^2+b^2 ≥ 2ab, developping (a-b)^2 is simple and straightforward
Isn't this fairly routine with some algebraic bash?
good for you?
Yeah I also thought this
You mean like mx plus a times nx plus b and you know a times b times b times b equals m right..why not do that??