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

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 \(7(F_{n+2})^{3}-(F_{n})^{3}-(F_{n+1})^{3}\).

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 \(a^{3}+b^{3}+c^{3}+abc=0\).

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

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TopNewestJust 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 \(F_{n+3}\).

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.

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

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– Ahmad Kharrat · 1 year, 5 months ago

n = 1.25 is a solution for question 1Log in to reply

– Shivam Jadhav · 1 year, 9 months 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 .Log in to reply

– Anupam Nayak · 1 year, 8 months ago

I could only solve the first two. My solution for the first question is about the same.Log in to reply

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 · 1 year, 9 months 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 · 1 year, 9 months ago

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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. – Aareyan Manzoor · 1 year, 9 months ago

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@Nihar Mahajan @Surya Prakash – Shivam Jadhav · 1 year, 9 months ago

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n = 1.25 is a solution for question 1 – Ahmad Kharrat · 1 year, 5 months ago

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When will be the result of rmo 2015 will be declared for rajasthan region?? – Aakash Khandelwal · 1 year, 9 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 · 1 year, 9 months ago

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– Aareyan Manzoor · 1 year, 9 months 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.Log in to reply

– Shrihari B · 1 year, 9 months 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.Log in to reply

– Aareyan Manzoor · 1 year, 9 months 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.Log in to reply

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

\[ \sqrt{ 3 + 2 \sqrt{2} } + \sqrt{ 3 - 2 \sqrt{2} } = 2 \] – Calvin Lin Staff · 1 year, 9 months ago

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These are questions from AMTI final 2014 for junior level isn't it ? – Shrihari B · 1 year, 9 months ago

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@Daniel Liu – Shivam Jadhav · 1 year, 9 months ago

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Q1 there are no such numbers – Satyajit Ghosh · 1 year, 9 months ago

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– Shivam Jadhav · 1 year, 9 months ago

Prove itLog in to reply

– Kushagra Sahni · 1 year, 9 months ago

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

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 · 1 year, 9 months ago

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