By 10:31 we take {x+y=16-z, and xy=168/z} then x,y are roots the polynomial f(t)=t^2-(16-z)+168/z. With discriminant is a square (x,y integer). But discriminant(f(t))=(16-z)^2-4*168/z. We can assume that |z| is minimal and divides 168. Thus |z|
Easier solution: move c to the other side, cube, combine with the second equation, we get (8-c)(ab+8c)=3.7.8, we just check few cases, for each case, we get a+b=something and ab=something. Then we get quickly the same answer.
Could have replaced (a+b) with (8-c) instead of x for easy and quick trial and error process. Thx for the creative ideas. Always enjoy watching the videos
expand cubic (gets sameformula as lemma) (b+c)(8-b)(8-c)=2^3*3*7 gives u the sum of factors of 16...find triplet = 16 and solve ...very few lines. Nice that author shows a variety of techniques through his videos
Could have used more detail at the end when showing which possible values of x yielded valid solutions. On the other hand, there was no need to take the second derivative of the cubic function. From the positive leading coefficient, you can see right away that x=16/3 is the local max and x=16 is the local min.
I tried to sketch the cubic function, and by finding the turning points, it shows that the inequality can only hold for integers >= 22. The rest is trying out positive factors of 168 that is >= 22.
I'm wrong but I dont know where: I did it by studying mod4 (since first I knew there was smthg with cubes mod 4 and didn't want too many cases): X-->X³: 0-->0 1-->1 2-->0 3-->1 The only way to get 0 mod 4 by summing 3 terms is: 4+0+0 3+1+0 2+2+0 2+1+1 We notice that if we cube each term (mod4) we get: 0+0+0 1+1+0 0+0+0 0+1+1 So only the first and third cases work, so a b c are even, I called a'=a/2 same with b and c. We get: a'+b'+c'=4 a'³+b'³+c'³=1 By studying again mod4 we see that there isnt any possibility to get 1 mod4 while satisfaying first line
3^3 = 27 is 3 mod 4. So the third case of 3+1+0 works out mod 4 and still needs to be considered. The only solution is of this form (15, 9, -16). (Also there is one more way to get 0 mod 4 from summing three terms: 3+3+2. But when you look at the cubes, this case doesn't come out to zero mod 4: 3^3 + 3^3 + 2^3 = 3+3+0=2 mod 4.)
By 10:31 we take {x+y=16-z, and xy=168/z} then x,y are roots the polynomial f(t)=t^2-(16-z)+168/z. With discriminant is a square (x,y integer). But discriminant(f(t))=(16-z)^2-4*168/z.
We can assume that |z| is minimal and divides 168. Thus |z|
Easier solution: move c to the other side, cube, combine with the second equation, we get (8-c)(ab+8c)=3.7.8, we just check few cases, for each case, we get a+b=something and ab=something. Then we get quickly the same answer.
how to combine with the second equation?
Could have replaced (a+b) with (8-c) instead of x for easy and quick trial and error process. Thx for the creative ideas. Always enjoy watching the videos
that is a lot of effort😘
expand cubic (gets sameformula as lemma) (b+c)(8-b)(8-c)=2^3*3*7 gives u the sum of factors of 16...find triplet = 16 and solve ...very few lines. Nice that author shows a variety of techniques through his videos
Could have used more detail at the end when showing which possible values of x yielded valid solutions. On the other hand, there was no need to take the second derivative of the cubic function. From the positive leading coefficient, you can see right away that x=16/3 is the local max and x=16 is the local min.
Great explanation!
I am a IOQM aspirant (Indian qualifier exam for IMO) and when i saw the qsn i was like, this shit from IMO brr? Then i started solving it...
Can you please explain the bit after 10:10 with the cubic inequality?
I tried to sketch the cubic function, and by finding the turning points, it shows that the inequality can only hold for integers >= 22. The rest is trying out positive factors of 168 that is >= 22.
Yeah I am completely lost on that after following along everything until that point.
thanks for sharing
I'm wrong but I dont know where:
I did it by studying mod4 (since first I knew there was smthg with cubes mod 4 and didn't want too many cases):
X-->X³:
0-->0
1-->1
2-->0
3-->1
The only way to get 0 mod 4 by summing 3 terms is:
4+0+0
3+1+0
2+2+0
2+1+1
We notice that if we cube each term (mod4) we get:
0+0+0
1+1+0
0+0+0
0+1+1
So only the first and third cases work, so a b c are even,
I called a'=a/2 same with b and c. We get:
a'+b'+c'=4
a'³+b'³+c'³=1
By studying again mod4 we see that there isnt any possibility to get 1 mod4 while satisfaying first line
It seems that 3³ has to be 3 mod 4
3^3 = 27 is 3 mod 4. So the third case of 3+1+0 works out mod 4 and still needs to be considered. The only solution is of this form (15, 9, -16).
(Also there is one more way to get 0 mod 4 from summing three terms: 3+3+2. But when you look at the cubes, this case doesn't come out to zero mod 4: 3^3 + 3^3 + 2^3 = 3+3+0=2 mod 4.)
I solved this one in my head simply by guessing possibilities and checking if they work. After a couple minutes, I came up with -16, 15 and 9.
Brilliant
This is engineering proof
always lmo