Representing binomial coefficients?
a , b , c are integers satisfying
a + b + c = a 2 + b 2 + c 2 = a 3 + b 3 + c 3 .
Are 0 and 1 the only possible values of a , b , c ?
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6 solutions
Yup, I've added the cubes part as a red herring.
Good follow up question. Lemme think about that.
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(4, 4, -5), and (-4, -4, 5) work for the x + y + z = x 3 + y 3 + z 3 ...
And (9, 15, -16) and (-9, -15, 16) if you want distinct integers. B-)
Also, obviously (x,0,-x).
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Yes, I think that's a nice follow-up question. Would you like to post a problem regarding this setup?
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@Pi Han Goh – Thanks for the suggestion, but i don't think i'm at that level yet. :-)
Of course, there are non-integer solutions. If 0 ≤ a ≤ ≈ 1 . 1 2 3 1 1 there are values for b and c that work. I don't know the significance of that upper number.
https://www.desmos.com/calculator/mgi3682y45
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Go ahead and post it as a problem then, maybe there's some clever way to tackle this without resorting to numerical methods?
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I would if I could find a closed form.
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@Jeremy Galvagni – For what it's worth, I sometimes post problems when I don't know how to "properly" solve them too. =)
Would be much more interesting without the integer constraint.
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True, but also quite a LOT more complicated... I tried to dig into that, to understand what's behind the 1.12311 bound that Jeremy Galvagni asserted below, but it got too messy too quickly. :-s
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I couldn't figure it out either. It would make a nice follow-up if there were a closed form.
Exactly, it didn't have -1 so I put no. It said 0 and 1 and that's wrong
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How is "only 0 and 1" wrong? The number -1 is not a solution of x^2 - x = 0...
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I don't see the relevance here. Care to elaborate on this?
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@Pi Han Goh – Huh... What exactly can you make of the comment i was replying to? Let me quote it: "Exactly, it didn't have -1 so I put no. It said 0 and 1 and that's wrong "
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@C . – I think you have misunderstood the question here.
The question claims that the possible values of a , b , c can be either 0 or 1 only.
Are you claiming that -1 can also be a possible value of at least one of a , b , c ?
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@Pi Han Goh – I think you should read again the whole thread, and this time maybe pay attention to who said what? -.-
When squared, all integers except for 0 and 1 result in a higher number.
This means if any of a , b or c are an integer besides 0 or 1 , then a 2 + b 2 + c 2 > a + b + c .
In the question, it is nowhere mentioned that the solution must be integer. It can be any number including decimal.
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It's mentioned in the first line: a,b,c are integers satisfying ...
Let a + b + c = a^2 + b^2 + c^ 2 = a^3 + b^3 + c^3 = x Since a+b+c = a^2 +b^2 + c^2 , we get a+ b+ c>=0 Case 1: a+ b+ c > 0
a + b + c = x
x = x^2 - 2 (ab + bc + ac)
=> x (x-1)/2 = ab + bc + ac
We know that ab + bc + ac <= a^2 + b^2 + c^2
Therefore x (x-1)/2 <= x
=> x <= 3
So 0 < a + b + c <=3
Now we can easily check from the given data
that For a + b + c = 1, 2 or 3 the solution for a,b
and c can only be 0 or 1
Case 2 :
a + b +c = 0
Since a+b+c = 0, we get a^3 + b^3 + c^3 = 3abc
But a+ b + c = a^3 + b^3 + c^3 which is given
So 3abc = 0
Let a = 0 , then we have 0 = 0 + b^2 + c^2 this
forces b and c to be 0 .
Thus 0, 1 are the only solutions for a,b and c
How do you get X = x^2 - 2 (ab + bc + ac)?
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X=a+b+c
X2=(a+b+c)(a+b+c)=
a2+ab+ac+ab+b2+bc+ac+bc+c2= a2+b2+c2+2ab+2bc+2ac
x=a2+b2+c2=x2-2(ab+bc+ac)
Edit: I've just seen the "integer" restriction. That renders this solution invalid. Still find it very interesting though.
There is a solution to this problem that is not really that hard to find.
The numbers can be:
-1, i, -i
Knowing that the number i is √-1, we can see how:
For a + b + c :
-1 +i -i = -1
For a² + b² + c² :
(-1)² + i² + (-i)² = 1 + (-1) + (-1) = 1 - 1 - 1 = -1
And finally, for a³ + b² + c² :
(-1)³ + i³ + (-i)³ = -1 + (-i) + (i) = -1 - i + i = -1
All 3 equalities add up to -1, thus fulfilling the requirements.
We know a ≤ a 2 for any integer a , and equality occurs only if a = 0 or if a = 1 . This shows
a + b + c ≤ a 2 + b 2 + c 2
for any integers a , b , c and equality occurs only if a , b , c ∈ { 0 , 1 } . For such values of a , b , c , above terms also equal to the cubic term obviously.
I don't know how to post a solution so I'm posting it as a comment. I wouldn't have bothered because they are pretty much all the same, except that mine is more general. The statement holds for any number of terms, so a 1, a 2, ..., a_n are all integers.
sum(a k^2) >= sum(a k) because a k^2 >= a k for all k. If a k is neither 0 nor 1 for some k, then a k^2 > a k, so sum(a k^2) > sum(a_k).
In other words, if a 1 + a 2 + ... + a n = a 1^2 + a 2^2 + ... + a n^2 (cubes are irrelevant), then every a_k has to be either 0 or 1.
a^2=a if a=0,1
Otherwise for any integer value of a except(0,1) a^2>a
Since if a>1 implies a^2>a if a<0 implies a(a-1)=negative ×negative >0 or a^2>a
So, as well as a^2+b^2+c^2>a+b+c
You didn't answer my question. Yes, 1 2 = 1 and 0 2 = 0 are indeed true, but you didn't justify why there's no other solutions (other than 0 and 1) for a , b , c .
The first equality implies (a^2 - a) + (b^2 - b) + (c^2 - c) = 0. The parenthesized expressions are non-negative integers, therefore all 0. The result follows immediately.
The equality involving cubes is not needed for the result to follow, but it does raise questions such as what can be said if only the two extremes are equal (values are only 0, 1, -1 ?).