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# INMO Practice questions

1. If $$n$$ is a positive integer . Then find values of $$n$$ such that $$\sqrt{n-1}+\sqrt{n+1}$$ is a rational number.

2. If a define a sequence of natural numbers such that $$F_{0}=F_{1}=1,F_{n}=F_{n-1}+F_{n-2}$$ for all $$n\geq2$$ . Then prove that $$F_{n+3}$$ divides $$7(F_{n+2})^{3}-(F_{n})^{3}-(F_{n+1})^{3}$$.

3. If $$x,y,z,a,b,c$$ are non-zero real numbers such that $$\frac{x^{2}(y+z)}{a^{3}}=\frac{y^{2}(x+z)}{b^{3}}=\frac{z^{2}(y+x)}{c^{3}}=\frac{xyz}{abc}=1$$ . Prove that $$a^{3}+b^{3}+c^{3}+abc=0$$.

4. If $$x,y,z>1$$ prove that $$\sum_{cyclic}^{x,y,z}\frac{x^{4}}{(y-1)^{2}}\geq48$$.

1 year, 1 month ago

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1. Note that both $$\sqrt{n-1}$$ and $$\sqrt{n+1}$$ are algebraic integers, so $$\sqrt{n-1}+\sqrt{n+1}$$ is also an algebraic integer. Since it is also rational, it follows that it is a rational integer. Thus, we just need to find when $$\sqrt{n-1}+\sqrt{n+1}=k$$ for some $$k\in\mathbb{Z}$$. Squaring gives $$2n+2\sqrt{n^2-1}=k^2$$ so $$\sqrt{n^2-1}$$ must be an integer, contradiction unless $$n=1$$. However, plugging that in gives$$\sqrt{2}$$ which is not integral so there are no solutions.

2. Just algebra: \begin{align*} 7F_{n+2}^3-F_{n+1}^3-F_n^3 &= 7F_{n+2}^3-(F_{n+1}+F_n)(F_{n+1}^2-F_{n+1}F_{n}+F_n^2)\\ &= F_{n+2}(7F_{n+2}^2-F_{n+1}^2+F_{n+1}F_n-F_n^2)\\ &= F_{n+2}(7(F_{n+1}+F_n)^2-F_{n+1}^2+F_{n+1}F_n-F_n^2)\\ &= F_{n+2}(6F_{n+1}^2+15F_{n+1}F_n+6F_n^2)\\ &= 3F_{n+2}(2F_{n+1}+F_n)(F_{n+1}+2F_n)\\ &= 3F_{n+2}F_{n+3}(F_{n+1}+2F_n) \end{align*} which is clearly divisible by $$F_{n+3}$$.

3. Note that $$a^3=x^2(y+z)$$. Multiplying the three cyclic variants of this equation gives $$a^3b^3c^3=x^2y^2z^2(x+y)(y+z)(z+x)$$. However, we also know $$abc=xyz$$ so plugging in and simplifying gives $$xyz=(x+y)(y+z)(z+x)\implies xyz+\sum\limits_{sym}x^2y=0$$. However, note that $$a^3+b^3+c^3+abc=xyz+\sum\limits_{cyc}x^2(y+z)=xyz+\sum\limits_{sym}x^2y=0$$ so we're done.

4. $$\displaystyle\sum_{cyc}\dfrac{a^4}{(b-1)^2}=\sum_{cyc}\dfrac{(a-1+1)^4}{(b-1)^2}\ge \sum_{cyc}\dfrac{(2\sqrt{a-1})^4}{(b-1)^2}=\sum_{cyc}\dfrac{16(a-1)^2}{(b-1)^2}\ge 48$$ where both inequalities are by AM-GM.

· 1 year, 1 month ago

n = 1.25 is a solution for question 1 · 9 months, 1 week ago

An easier way to solve question number 2. Is by writing in form of $=8F^{3}_{n+2}-F^{3}_{n}-(F^{3}_{n+1}+F^{3}_{n+2})$ And then expand we get in one bracket $$(2F_{n+2}-F_{n})=(F_{n+2}+F_{n+2}-F_{n})=(F_{n+2}+F_{n+1})=(F_{n+3})$$ and other bracket as $$(F_{n+1}+F_{n+2})=(F_{n+3})$$ and hence we take $$F_{n+3}$$ as overall common and the result gets proved . · 1 year, 1 month ago

I could only solve the first two. My solution for the first question is about the same. · 1 year ago

In question 4th X^2 +4>= 4x thus x^4>=16(x-1)^2 Dividing by (y-1)^2 and adding for all x,y,z Then apply AM.GM and get the result · 1 year, 1 month ago

In 3 question put values of a^3 b^3 c^3 and abc from the question and it can be easily proved. · 1 year, 1 month ago

q1) write as one radical $\sqrt{2n+2\sqrt{n^2-1}}$. let this be rational number k. then $2n+2\sqrt{n^2-1}=k^2$ the LHS is rational, so RHS must be rational.this implies n^2-1 is a square of a rational number. as n is an integer, that rational number Must also be an integer. this implies $n^2-1=a\Longrightarrow (n-a)(n+a)=1=1*1=-1*-1$ $n=1\quad or\quad -1(n/a)$ at n=1, the expression is not a perfect square, so no positive integral solution. · 1 year, 1 month ago

@Nihar Mahajan @Surya Prakash · 1 year, 1 month ago

n = 1.25 is a solution for question 1 · 9 months, 1 week ago

When will be the result of rmo 2015 will be declared for rajasthan region?? · 1 year, 1 month ago

For question 1 Since the sum of two square roots is rational implies both are rational. But n is integer which forces each term to be an integer We can just say that both n-1 and n+1 are perfect squares and the difference between these two perfect squares is 2 which does not yield any solution. · 1 year, 1 month ago

"the sum of two square roots is rational implies both are rational", this is not justified, marks would not have been given. justify it please. · 1 year, 1 month ago

I don't think any justification is required for such a trivial statement. Still if u want it is as follows . The two square roots could be (rational,rational),(rational,irrational),(irrational,irrational) rational plus irrational can't give rational. irrational plus rational can give rational only if both cancel each other which means one of the square roots is negative which is again ruled out So u r left with the only possibility of both being rational. · 1 year, 1 month ago

yes,that is right, but one must add these to math olympiads. or marks might be deducted. but why cant we have something like $2.1357843257...+2.8643...=5$. justifying this the main challenge. · 1 year, 1 month ago

Yup, that is a good point. Especially since that observation is the crux of the problem, it is crucial to prove it, or state how one can prove it. The idea of "only if both cancel out" is circular. Essentially, what you are saying is: The reason why there are no non-perfect square solutions to $$\sqrt{a} + \sqrt{b} = n$$, is because there are there are no non-perfect square solutions to $$\sqrt{a} = n - \sqrt{n}$$. And vice versa. Thus, nothing has been shown just as yet.

Another example where the terms are explicitly square roots (just not of integers) is:

$\sqrt{ 3 + 2 \sqrt{2} } + \sqrt{ 3 - 2 \sqrt{2} } = 2$ Staff · 1 year, 1 month ago

These are questions from AMTI final 2014 for junior level isn't it ? · 1 year, 1 month ago

@Daniel Liu · 1 year, 1 month ago

Q1 there are no such numbers · 1 year, 1 month ago

Prove it · 1 year, 1 month ago

Why does the question ask for real numbers? Every positive integer n will make the expression a real number. · 1 year, 1 month ago