The Answer Is At Least 1
How many positive integers m are there, such that for some triple of positive integers ( a , b , c ) , we have
m = b a + c b + a c ?
For example, if a = b = c , then we get the value 1 1 + 1 1 + 1 1 = 3 .
Hence, the answer is "at least 1", which applies to all 4 options. You need to figure out which is correct.
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4 solutions
This question is easily misunderstood, as seen by the other solutions. I've rephrased it, to try and make the distinction clearer. The answer is also somewhat easily guessed.
For these kind of problems, to prove "infintely many", you just need to find a family of solutions, as opposed to trying to classify all possible solutions. In fact, classifying all solutions is an open problem.
(spoiler below)
Hint: m = k 2 + 5 is a family of solutions which work.
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I see that this problem has already dropped down to Level TWO!
It's definitely the HARDEST Level 2 problem I've ever run across here in Brilliant! I can't hardly wait to see the solution even for limited values of m. I can see now that there's an infinity of m, but proving it seems really elusive.
Okay, here it is, finally:
For any integer n , let
x = n 2 − n + 1
y = 2
x = n 2 + n + 1
a = x 2 y
b = y 2 z
c = z 2 x
Then it all reduces to
m = n 2 + 5
Thanks for the hint, Calvin! This is my favorite really difficult Level 2 problem.
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@Michael Mendrin How do you get this? I mean what was your motivation behind this?
@Calvin Lin Please help me out.
Although its a very very late reply.
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@Ankit Kumar Jain – This is a construction problem and you have to make some guesses as to what works and what doesn't.
This is how I engineered the problem.
First, I was studying the equation N x y z = x 3 + y 3 + z 3 , which is a somewhat common appearance in olympiad problems. This has non-trivial solutions for infinitely many N-values, of which Michael found one family with the hint of N = n 2 + 5 . There are other families, and classifying them seems to be a huge pain. There are papers written on this equation, which you should be able to find by doing a search.
After that, I used the substitution that Michael stated, we get the problem. I took the substitution from somewhere else (possibly a question about inequalities). I remembered it because of this nice forms that it gave.
Omg, I misread the problem but still got it right. I thought it asked for the number of triplets...
How did you assume something like x,y with n ? My point is I want to learn how you thought to approach because it seems too straight-forward.
a= 2^x, b= 2^2x , c= 2^3x
a = 2 n , b = 2 n + 1 , c = 2 n + 2
Doesn't that only give 2 1 + 2 1 + 4 = 5 always? Why would the answer be "infinitely many m"?
I'm swear that I answered infinitely many but why I comes to not choose that answer by brilliant. Please makes it clear -_-
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Thanks. We will look into this. It seems like some people have been getting this issue, possibly related to the internet connection.
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I CLICKED INFINITELY MANY TOO BUT IT SHIFTED TO 2 ....HOW
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@Ashutosh Mahapatra – We have fixed a bug in which answers were not properly recorded. Please let me know if you still see any issues going forward.
Can you explain how you got to that solution?
you can have many solutions. The easiest way to do it is a=b=c. You can choose a to be 1,2,3,..................,9999999999,....infinity
we can see that if a, b, c are in ratio 1:2:4 then it follows...... Since we have many such triplets Therefore INFINITELY
Once a = b = c it satisfies the condition.
It does satisfy the condition, but that only generates 1 value of m. (and hence the title "The answer is at least 1", since that is the most obvious choice)
Why are there "infinitely many positive integers m" which can be expressed in that way?
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correct...only single value of 'm' would come when a=b=c... since the question is not asking for the no. of values of a,b or c instead its asking for no. of values of 'm', so in my opinion also it should be at least 1, as its most correct in this context..
(Solution from comments)
For any integer n , let
x = n 2 − n + 1
y = 2
z = n 2 + n + 1
a = x 2 y
b = y 2 z
c = z 2 x
Then it all reduces to
m = n 2 + 5
This is a Level 3 problem? I guessed at "Infinitely many" only because I couldn't even get a start on proving that there are infinitely many. I'm still working on that proof. We know that there are more than 2. Very tough problem! For me anyway.
The best I can do now is to convert this to a 2nd order Diophantine equation, where m is arbitrary. But I'm not sure if a solution always exists for arbitrary m, which is why I don't know if infinitely many m are possible.