8 votes
Accepted

Two element subset of $\{1,2,\dots,100\}$ with sum of elements being a square

You have the right idea re: checking $2$-element subsets. However, instead, subdivide the elements of $\{1,2,\ldots,100\}$ into the $50$ disjoint $2$-element subsets of $\{1,99\}$, $\{2,98\}$, $\ldots\...
John Omielan's user avatar
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7 votes
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If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$

Some thoughts. By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a^2 + 4bc}} \right)^2\cdot \sum_{\mathrm{cyc}} (a^2 + 4bc)(4b + 4c - bc + 4ab + 4ac)^3 \\ \...
River Li's user avatar
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7 votes
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counting sequences of elements of the set {1,2,3,4} with given property

Disclaimer This answer skips some details, but hopefully it's sufficient to guide you and let you research the details yourself. The Answer $$ x_{4+2m}= \left(\frac{5-3\sqrt{5}}{10}\right) \left(\frac{...
Rezha Adrian Tanuharja's user avatar
5 votes

Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.

A simple long division method yields $p(x)=ax^5+bx^4+1=(x^2-x-1)((ax^3)+(a+b)x^2+(2a+b)x+(3a+2b))+\ (5a+3b)x+(3a+2b+1)$ as we know $(x^2-x-1)$ is factor of $p(x)$, The remainder $(5a+3b)x+(3a+2b+1)$ ...
Dheeraj Gujrathi's user avatar
5 votes

If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$

Some thoughts. By AM-GM, it suffices to prove that $$\frac{2}{\frac{a^2 + 4bc}{2 + bc/2} + (2 + bc/2)} + \frac{2}{\frac{b^2 + 4ca}{2 + ca/2} + (2 + ca/2)} + \frac{2}{\frac{c^2 + 4ab}{2 + ab/2} + (2 + ...
River Li's user avatar
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4 votes
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Proving the existence of integers $a, b, c$ such that $\left\lvert a + b\sqrt{2} + c\sqrt{3}\right\rvert < 10^{-5}$

DISCUSSION : There are a couple of Issues with your nice approach. When those Issues are rectified , the Solution will work out. (1) The range you have taken is $10^3$ , though the Correct range is $...
Prem's user avatar
  • 7,458
4 votes

If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$

Remark 1: This is my third proof (the idea) which is better than my two old proofs. Remark 2: Some years ago, I used a similar idea for the problem: Let $a, b, c > 0$ with $a + b + c = 3$. Prove ...
River Li's user avatar
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4 votes

If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$

Disclaimer: Not a full solution If one of $a, b, c=0$: WLOG let $a=0$. Since $ab+bc+ca>0$ we know that $b, c\neq0$. $$\begin{align} & \frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ca}}+\frac{1}{\...
IraeVid's user avatar
  • 2,940
4 votes
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Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.

Your method is on-target. You just need to reduce the quintic until both the equations are of same degree to compare coefficients. I will show an example with highest degree of $2$ (reduction to ...
Dstarred's user avatar
  • 2,352
4 votes

Proving $\sum_{i=0}^{n}\sum_{j=0}^{n}\binom{i+j}{i}\binom{n-i}{j}\binom{n-j}{i}\ =\ \sum_{k=0}^{n}\binom{2k}{k} $

The following standard identities are used in the answer: $$\begin{align} \frac{1}{1-x} &= \sum\limits_n x^n, \\ (x+y)^s &= \sum\limits_{i+j=s} \binom{i+j}{i} x^i y^j, \\ \frac{1}{(1-x)^{k+1}} ...
Oleksandr  Kulkov's user avatar
3 votes
Accepted

If $1! \cdot 2! \cdot 3! \cdot\cdots\cdot 12! = m! \, n^2$, then what is $m$?

From this step: $= (1!)^2 \cdot 2 \cdot (3!)^2 \cdot 4 \cdot (5!)^2 \cdot 6 \cdot \ldots \cdot (11!)^2 \cdot 12$ Group factorials and separate even numbers: $= (1! \cdot 2! \cdot 3! \cdot \ldots \cdot ...
FishDrowned's user avatar
3 votes
Accepted

IMO 1987/P1 - Combinatoric approach

Your claim that $p_n(k)=\binom nk(n-k-1)!$ is not correct. In the case where $k=0$, your formula would imply that there are $(n-1)!$ permutations with zero fixed points, so $(n-1)!$ derangements of $n$...
Mike Earnest's user avatar
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2 votes

counting sequences of elements of the set {1,2,3,4} with given property

Here's a recurrence I got. Let $f_n$ be the number of ways to have such a sequence of length $n$ with $a_n = 2$. Case 1: $a_{n-1}=1$ $a_{n-2}$ must be 2, this accounts for $f_{n-2}$ ways. Case 2: $a_{...
Shreya Mundhada's user avatar
2 votes

Monochromatic $4$-Cycle in Bipartite Complete Graph

Let one part be $\{\,v_1, v_2, \ldots, v_n\,\}$ and another part be $\{\,u_1, u_2, \ldots, u_n\,\}$. Let's show that for $n = 5$ the graph always has a monochromatic $4$-cycle. Suppose the opposite. ...
Smylic's user avatar
  • 6,390
2 votes

Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.

Notice that $\frac{1\pm\sqrt{5}}{2}$ are the roots of $p(x)$ as they are the roots of $x^2-x-1$. Hence, we would have two equations: $$p\left(\frac{1+\sqrt{5}}{2}\right) = a\left(\frac{1+\sqrt{5}}{2}\...
Yathiraj Sharma's user avatar
2 votes

A polynomial's coefficients are non-negative integers, $f(1)=6$ and $f(7)=3438$. What is the value of $f(3)$?

Suppose that $p$ is a positive integer, $M\in\Bbb N$ and $$M=\sum_{k=0}^na_kp^k=\sum_{k=0}^nb_kp^k$$ such that $$a_k,b_k\in\{0,1,...,p-1\}\tag1$$ Then $$\sum_{k=0}^n(a_k-b_k)p^k=0\tag2.$$ From $(2)$, ...
Bob Dobbs's user avatar
  • 8,614
1 vote

Prove that $\sum_{i=0}^{\lfloor n/2 \rfloor }\sum_{j=0}^{\lfloor n/2 \rfloor-i}{n \choose 2i}{n-2i \choose 2j}2^{n-2i-2j}=4^{n-1}+2^{n-1}$

Supposing we start from $$\sum_{p=0}^n \frac{1}{2} (1+(-1)^p) \sum_{q=0}^{n-p} \frac{1}{2} (1+(-1)^q) {n\choose p} {n-p\choose q} 2^{n-p-q} = 4^{n-1} + 2^{n-1}.$$ This is $$\sum_{p=0}^n (1+(-1)^p) {n\...
Marko Riedel's user avatar
  • 60.4k
1 vote
Accepted

A polynomial's coefficients are non-negative integers, $f(1)=6$ and $f(7)=3438$. What is the value of $f(3)$?

So regrading the question you have in their solution : Let me explain in such a way : a 3 digit number in base 10 (what we use currently : the decimal representation) is expressed as $a 10^0 + b 10^1 +...
Ham Lemon's user avatar
  • 537
1 vote
Accepted

German MO combinatorics problem 1995

Just to clarify user 10354138 answer beacause there might be a small mistake: $$\frac{k}{2n-k+2}+\frac{2n-k+1}{k+1}\ge 2\cdot \frac{n+1}{n+2}$$ $$\iff \frac{k}{2n-k+2}+\frac{2n-k+1}{k+1}-\frac{2(n+1)}{...
maxime's user avatar
  • 26

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