Hi! This week's set consists in three problems in algebra that seem nice. I'm using the word "seem" because I've been working on them and haven't been able to solve them. Anyway, enjoy the problems, and, if you find them nice and/or fun, please reshare the note, so more people will be able to see the problems and try to tackle these hard problems. Just as a reference, problem 1 was taken from the Colombian book "Un recorrido por el álgebra" and problems 2 and 3 were taken from the first Costa Rican IMO TST 2014.

Find all functions \(f:\mathbb{Z}^+\rightarrow\mathbb{R}\) such that \[f(n+m)+f(n-m)=f(3n)\] for every \(n,m\in{\mathbb{Z}^+}, n>m\).

Find all real solutions \((x_1,...,x_n)\) of the equation \[(1-x_1)^2+(x_1-x_2)^2+...+(x_{n-1}-x_n)^2+x_n^2=\dfrac{1}{n+1}\]

Let \(x, y\) be real numbers such that \(y>0\) and \(7x^2+3xy+3y^2=1\). Show that \[\dfrac{x^2+y^2}{y}\ge{\dfrac{1}{2}}\]

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TopNewestFor the second one:

Using the Cauchy-Schwarz Inequality we can write, \( \displaystyle \begin{array} \\ \bigg( (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \bigg) \bigg( 1^2 + 1^2 + \dotsb + 1^2 + 1^2 \bigg) \geq & \\ \bigg( (1-x_1) + (x_1 - x_2 ) +\dotsb + (x_{n-1} - x_n)+ x_n \bigg)^2 & \\ \end{array} \)

\( \displaystyle \Rightarrow \Big( (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \Big) (n+1) \geq ( 1)^2 \) \(\displaystyle \Rightarrow (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \geq \frac{1}{n+1} \)

But it is given that the equality condition holds true. Therefore, from the condition of equality in Cauchy-Schwarz, we obtain,

\( \displaystyle \frac{1-x_1}{1} = \frac{x_1 - x_2}{1} = \dots = \frac{x_{n-1} - x_n}{1} = \frac{x_n}{1} \) \(\displaystyle \Rightarrow x_{n-1} = 2x_n \text{ , } x_{n-2} = 3x_n ,\dotsc, \text{ } x_2 = (n-1)x_n \text{ and } x_1 = nx_n \).

Also, \(\displaystyle 1 = x_1 + x_n \Rightarrow (n+1)x_n = 1 \Rightarrow x_n = \frac{1}{n+1} \).

Hence, \(\displaystyle x_i = \frac{n+1-i}{n+1} \text{ ; } i = 1,2,3, \dotsc , n-1, n \) – Sudeep Salgia · 3 years ago

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– Shaun Loong · 3 years ago

May I know how you arrive at a conclusion for \(x_{i}\)?Log in to reply

– Samuraiwarm Tsunayoshi · 3 years ago

From third last line, \(x_{i} = (n+1-i)x_{n}\) and \(x_{n} = \frac{1}{n+1}\).Log in to reply

– Shaun Loong · 3 years ago

Thanks, I see it now :)Log in to reply

For no.1 I have a weird start, this might helps.

Substituting \(m \rightarrow 1, 2, 3, ..., n-1\) we get

\(f(2n-i) + f(i) = f(3n) = f(2n-j) + f(j)\) for every \(i,j = 1, 2, ..., n-1\)

We see that if \(a+b = c+d\), then \(f(a)+f(b)=f(c)+f(d)\) for positive integers \(a,b,c,d\).

I have no idea to do anything right meow. I suck at this thing XD – Samuraiwarm Tsunayoshi · 3 years ago

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I know of a "crude" solution to 3, which only uses basic knowledge of conic sections. It has a nice interpretation that way. – Calvin Lin Staff · 3 years ago

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– José Marín Guzmán · 3 years ago

I'm afraid I don't know much about conic sections, but share the solution and I'll try to look up the stuff I don't understand :). Besides, it would be useful for other people who may know about conic sections.Log in to reply

Hence, it follows that for any point on the ellipse which satisfy \( y \geq 0 \), we have

\[ 2 x^2 + 2y^2 - y \geq 0 \Rightarrow \frac{ x^2+y^2} { y} \geq \frac{1}{2}. \]

The tangential point is the equality case. – Calvin Lin Staff · 3 years ago

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– Abhinav Raichur · 3 years ago

awesome man!!........... nice approach....... thanks:)Log in to reply

For question \(2\), does the solution only occur when \(n=0\) with \(x_{0}=1\)? – Shaun Loong · 3 years ago

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Feel free to post ideas and solutions! :D – José Marín Guzmán · 3 years ago

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