INMO Practice questions

  1. If nn is a positive integer . Then find values of nn such that n1+n+1\sqrt{n-1}+\sqrt{n+1} is a rational number.

  2. If a define a sequence of natural numbers such that F0=F1=1,Fn=Fn1+Fn2F_{0}=F_{1}=1,F_{n}=F_{n-1}+F_{n-2} for all n2n\geq2 . Then prove that Fn+3F_{n+3} divides 7(Fn+2)3(Fn)3(Fn+1)37(F_{n+2})^{3}-(F_{n})^{3}-(F_{n+1})^{3}.

  3. If x,y,z,a,b,cx,y,z,a,b,c are non-zero real numbers such that x2(y+z)a3=y2(x+z)b3=z2(y+x)c3=xyzabc=1\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 a3+b3+c3+abc=0a^{3}+b^{3}+c^{3}+abc=0.

  4. If x,y,z>1x,y,z>1 prove that cyclicx,y,zx4(y1)248\sum_{cyclic}^{x,y,z}\frac{x^{4}}{(y-1)^{2}}\geq48.

Note by Shivam Jadhav
4 years ago

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  1. Note that both n1\sqrt{n-1} and n+1\sqrt{n+1} are algebraic integers, so n1+n+1\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 n1+n+1=k\sqrt{n-1}+\sqrt{n+1}=k for some kZk\in\mathbb{Z}. Squaring gives 2n+2n21=k22n+2\sqrt{n^2-1}=k^2 so n21\sqrt{n^2-1} must be an integer, contradiction unless n=1n=1. However, plugging that in gives2\sqrt{2} which is not integral so there are no solutions.

  2. Just algebra: 7Fn+23Fn+13Fn3=7Fn+23(Fn+1+Fn)(Fn+12Fn+1Fn+Fn2)=Fn+2(7Fn+22Fn+12+Fn+1FnFn2)=Fn+2(7(Fn+1+Fn)2Fn+12+Fn+1FnFn2)=Fn+2(6Fn+12+15Fn+1Fn+6Fn2)=3Fn+2(2Fn+1+Fn)(Fn+1+2Fn)=3Fn+2Fn+3(Fn+1+2Fn) \begin{aligned} 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{aligned} which is clearly divisible by Fn+3F_{n+3}.

  3. Note that a3=x2(y+z)a^3=x^2(y+z). Multiplying the three cyclic variants of this equation gives a3b3c3=x2y2z2(x+y)(y+z)(z+x)a^3b^3c^3=x^2y^2z^2(x+y)(y+z)(z+x). However, we also know abc=xyzabc=xyz so plugging in and simplifying gives xyz=(x+y)(y+z)(z+x)    xyz+symx2y=0xyz=(x+y)(y+z)(z+x)\implies xyz+\sum\limits_{sym}x^2y=0. However, note that a3+b3+c3+abc=xyz+cycx2(y+z)=xyz+symx2y=0a^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. cyca4(b1)2=cyc(a1+1)4(b1)2cyc(2a1)4(b1)2=cyc16(a1)2(b1)248\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.

Daniel Liu - 3 years, 12 months ago

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An easier way to solve question number 2. Is by writing in form of =8Fn+23Fn3(Fn+13+Fn+23)=8F^{3}_{n+2}-F^{3}_{n}-(F^{3}_{n+1}+F^{3}_{n+2}) And then expand we get in one bracket (2Fn+2Fn)=(Fn+2+Fn+2Fn)=(Fn+2+Fn+1)=(Fn+3)(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 (Fn+1+Fn+2)=(Fn+3)(F_{n+1}+F_{n+2})=(F_{n+3}) and hence we take Fn+3F_{n+3} as overall common and the result gets proved .

Shivam Jadhav - 3 years, 12 months ago

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n = 1.25 is a solution for question 1

Ahmad Kharrat - 3 years, 7 months ago

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I could only solve the first two. My solution for the first question is about the same.

Anupam Nayak - 3 years, 11 months ago

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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

Shubham Gupta - 4 years ago

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In 3 question put values of a^3 b^3 c^3 and abc from the question and it can be easily proved.

Shubham Gupta - 4 years ago

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q1) write as one radical 2n+2n21\sqrt{2n+2\sqrt{n^2-1}}. let this be rational number k. then 2n+2n21=k22n+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 n21=a(na)(n+a)=1=11=11n^2-1=a\Longrightarrow (n-a)(n+a)=1=1*1=-1*-1 n=1or1(n/a)n=1\quad or\quad -1(n/a) at n=1, the expression is not a perfect square, so no positive integral solution.

Aareyan Manzoor - 3 years, 12 months ago

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These are questions from AMTI final 2014 for junior level isn't it ?

Shrihari B - 3 years, 12 months ago

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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.

Shrihari B - 3 years, 12 months ago

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"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.

Aareyan Manzoor - 3 years, 12 months ago

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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.

Shrihari B - 3 years, 12 months ago

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@Shrihari B 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...=52.1357843257...+2.8643...=5. justifying this the main challenge.

Aareyan Manzoor - 3 years, 12 months ago

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@Aareyan Manzoor 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 a+b=n \sqrt{a} + \sqrt{b} = n , is because there are there are no non-perfect square solutions to a=nn \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:

3+22+322=2 \sqrt{ 3 + 2 \sqrt{2} } + \sqrt{ 3 - 2 \sqrt{2} } = 2

Calvin Lin Staff - 3 years, 11 months ago

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When will be the result of rmo 2015 will be declared for rajasthan region??

Aakash Khandelwal - 3 years, 12 months ago

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n = 1.25 is a solution for question 1

Ahmad Kharrat - 3 years, 7 months ago

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Q1 there are no such numbers

Satyajit Ghosh - 4 years ago

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Prove it

Shivam Jadhav - 4 years ago

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Why does the question ask for real numbers? Every positive integer n will make the expression a real number.

Kushagra Sahni - 4 years ago

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If p,q,r are positive and in AP.Then show that quadratic equation px^2+qx+r=0 are real for |p/r-7|greater than or equal to 4√3

Pushkar Sharma - 4 years ago

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