399. Algebra (I).

by Virgil Nicula, Sep 17, 2014, 3:51 PM

PP1 (Israel Diaz). Solve over $\mathbb R$ the system $\left\{\begin{array}{c}
x+y+z=3\\\\
\frac 1x+\frac 1y+\frac 1z=\frac 5{12}\\\\
x^3+y^3+z^3=45\end{array}\right\|$ .

Proof. Is well-known that $\left\{\begin{array}{c}
x+y+z=s_1\\\\
xy+yz+zx=s_2\\\\
xyz=s_3\end{array}\right\|\iff$ $t^3-s_1t^2+s_2t-s_3=0\odot\begin{array}{ccc}
\nearrow & x & \searrow\\\\
\rightarrow & y & \rightarrow\\\\
\searrow & z & \nearrow\end{array}\odot\ (*)$ . Let $S_n=x^n+y^n+z^n$ , where $n\in\mathbb N$ . Prove easily that

$S_{n+3}=s_1S_{n+2}-s_2S_{n+1}+s_3S_n$ , where $n\in\mathbb N$ and $\left\{\begin{array}{c}
S_0=3\\\\
S_1=s_1\\\\
S_2=s_1^2-2s_2\end{array}\right\|$ . For example, $S_3=s_1^3-3s_1s_2+3s_3$ . The our system becomes $\left\{\begin{array}{c}
s_1=3\\\\
\frac {s_2}5=\frac {s_3}{12}=k\\\\
S_3=45\end{array}\right\|$

and the equation $(*)$ becomes $t^3-3t^2+5kt-12k=0\odot\begin{array}{ccc}
\nearrow & x & \searrow\\\\
\rightarrow & y & \rightarrow\\\\
\searrow & z & \nearrow\end{array}\odot$ . Therefore, $S_3=45\iff$ $27-45k+36k=45\iff k=-2$ , i.e. $\left\{\begin{array}{c}
s_1=3\\\\
s_2=-10\\\\
s_3=-24\end{array}\right\|$ ,

Thus, $\{x,y,z\}$ are the roots of the equation $t^3-3t^2-10t+24=0$ , i.e. $\boxed{\{x,y,z\}=\{-3,2,4\}}$ (3!=6 solutions).



PP2. Find the maximum value of the parameter $a\in\mathbb R$ so that $(\forall )\ x>0\ ,\ x^3-ax+2\ge 0$ .

Proof. $(\forall )\ x>0\ ,\ x^3-ax+2\ge 0\iff$ $(\forall )\ x>0\ ,\ a\le x^2+\frac 2x\iff$ $a\le \min_{x>0}\left(x^2+\frac 2x\right)=3$ ,

i.e. $\max a=3$ because for any $x>0\ ,\ x^2+\frac 2x=x^2+\frac 1x +\frac 1x\ge 3\sqrt [3]{x^2\cdot\frac 1x\cdot\frac 1x}=3\implies a=3$ .


An easy extension. Find the maximum value of $a\in\mathbb R$ so that $(\forall )\ x>0\ ,\ x^{p+1}-ax+p\ge 0$ , where $p\in\mathbb N\ ,\ p\ge 2$ . This property means

in the analytical geometry that the line $y=(p+1)x-p$ is tangent to the graph $\mathbb G_f$ of the function $f(x)=x^{p+1}\ ,\ x\ge 0$ and pass through the point $P(0,-p)$ .



PP3 (Israel Diaz). Prove that $\{a,b,c\}\subset\mathbb R^*_+\ ,\ abc=1\ \implies\ \frac {a}{(a+1)(b+1)}+$ $\frac b{(b+1)(c+1)}+\frac c{(c+1)(a+1)}\ge \frac 34$ .

Proof. $(*)\ \sum_{\mathrm{cyc}}\frac {a}{(a+1)(b+1)}\ge \frac 34\iff$ $4\sum_{\mathrm{cyc}} a(c+1)\ge 3(a+1)(b+1)(c+1)\iff$ $4\left(s_1+s_2\right)\ge 3\left(1+s_1+s_2+s_3\right)\iff$ $s_1+s_2\ge 3\left(1+s_3\right)$ , where

$\left\{\begin{array}{c}
s_1=a+b+c\\\\
s_2=ab+bc+ca\\\\
s_3=abc=1\end{array}\right\|$ . Thus, $(*)\iff$ $s_1+s_2\ge 6$ what is truly because $s_1+s_2=a+b+c+ab+bc+ca=$ $a+b+c+\frac 1c+\frac 1a+\frac 1b=$ $\left(a+\frac 1a\right)+\left(b+\frac 1b\right)+$

$\left(c+\frac 1c\right)\ge 2+2+2=6$ with equality iff $a=b=c=1$ .



PP4. Find the real pairs $(x,y)$ so that $\sqrt x+\sqrt y+\sqrt {x+y-4}=2+\sqrt {xy}$ .

Proof. $\sqrt x+\sqrt y+\sqrt {x+y-4}=2+\sqrt {xy}\iff$ $\boxed{\sqrt {x+y-4}-1=\left(\sqrt x-1\right)\left(\sqrt y-1\right)}\ (*)$ . I"ll use the substitutions $\left\{\begin{array}{c}
x=(a+1)^2\\\\
y=(b+1)^2\end{array}\right\|$ , where $\{a,b\}\subset [-1,\infty )$ .

Therefore, the relation $(*)$ becomes $\sqrt {(a+1)^2+(b+1)^2-4}=1+ab\iff$ $(a+1)^2+(b+1)^2=4+(1+ab)^2\iff$ $(a+1)^2-2^2=(1+ab)^2-(b+1)^2\iff$

$(a-1)(a+3)=b(a-1)[(a+1)b+2]\iff$ $(a-1)\left[(a+1)b^2+2b-(a+3)\right]=0\iff$ $(a-1)(b-1)[(a+1)b+(a+3)]=0\iff$

$a=1\ \vee\ b=1\ \vee\ ab+a+b+3=0$ . Since $ab+a+b+3=(a+1)(b+1)+2\ge 2$ obtain that the relation $(*)$ is equivalently with $a=1\ \vee\ b=1\iff$

$\boxed{x=4\ \vee\ y=4}$ . Thus, all required pairs are $(4,\lambda )\ \vee\ (\lambda ,4)$ , where $\lambda\ge 0$ .



PP5 (Israel Diaz). Find the set $S$ of $n\in\mathbb N^*$ so that $(\forall )\ x\in\mathbb R\ ,\ \cos 2x=\cos^nx-\sin^nx$ .

Proof 1. Suppose w.l.o.g. $x\in\left(0,\frac {\pi}{2}\right)$ and $x\ne\frac {\pi}{4}$ . Thus, $\cos 2x=\cos^nx-\sin^nx\iff$ $\frac {\cos 2x}{1-\tan x}=\cos ^n x\cdot\frac {1-\tan^nx}{1-\tan x}\iff$ $\frac {1+\tan x}{1+\tan^2x}=\cos^nx\cdot\frac {1-\tan^nx}{1-\tan x}$ .

For $x\rightarrow\frac {\pi}{4}$ obtain that $1=\left(\frac {\sqrt 2}{2}\right)^n\cdot n\iff$ $\left(\sqrt 2\right)^n=n\iff$ $2^n=n^2\iff$ $\boxed{S=\{2,4\}}$ . I used $\cos 2x=\frac {1-\tan^2x}{1+\tan^2x}$ and $\lim_{x\to 1}\frac {x^a-1}{x-1}=a$ , where $a\ne 0$ .

Proof 2. Prove easily that $\{2,4\}\subset S$ and $\{1,3\}\subset \overline S$ . Suppose that $n\ge 5$ . In this case $(\forall )\ x\ne 0\ ,\ \cos 2x=$ $\cos^nx-\sin^nx\iff$ $\frac {1-\cos 2x}{x^2}=\frac {1-\cos^nx}{x^2}+\frac {\sin^nx}{x^2}\iff$

$4\cdot\frac {1-\cos 2x}{(2x)^2}=$ $\frac {1-\cos^nx}{1-\cos x}\cdot\frac {1-\cos x}{x^2}+\left(\frac {\sin x}{x}\right)^2\cdot\sin^{n-2}x$ . For $x\rightarrow 0$ and $\lim_{x\to 0}\frac {\sin x}{x}=1\ ,\ \lim_{x\to 0}\frac {1-\cos x}{x^2}=$ $\frac 12\ ,\ \lim_{x\to 1}\frac {x^a-1}{x-1}=a\implies$ $2=\frac n2$ abs. So $S=\{2,4\}$ .


PP6 (Israel Diaz). Prove that $\{a,b,c\}\subset \mathbb R^*_+$ and $\frac 1a+\frac 1b+\frac 1c=1\ \implies\ a^2+$ $b^2+c^2\ge 2(a+b+c)+9$ .

Proof. Observe that $\min\{a,b,c\}\ge 1$ and $(a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)\ge 9\implies$ $ \boxed{a+b+c\ge 9}\ (1)\ \implies$ $3\left[(a-1)^2+(b-1)^2+(c-1)^2\right]\ge $

$\left[(a-1)+(b-1)+(c-1)\right]^2=$ $\left[(a+b+c)-3\right]^2\stackrel{(1)}{\ge}(9-3)^2=36 \implies$ $\left[(a-1)^2+(b-1)^2+(c-1)^2\right]\ge 12\implies$ $a^2+$ $b^2+c^2\ge 2(a+b+c)+9$ .



PP7. Find the minimum value of the function $f(x)=\frac {\sin 3x}{\sin^3x}-\frac {\cos 3x}{\cos^3x}\ ,\ x\not\in\frac {\pi}{2}\cdot\mathbb Z$ .

Proof 1. $f(x)=\frac {3-4\sin^2x}{\sin^2x}-\frac {4\cos^2x-3}{\cos^2x}=$ $\frac {3\left(\sin^2x+\cos^2x\right)-4\sin^2x}{\sin^2x}-\frac {4\cos^2x-3\left(\sin^2x+\cos^2x\right)}{\cos^2x}=$ $\frac {3\cos^2x-\sin^2x}{\sin^2x}-\frac {\cos^2x-3\sin^2x}{\cos^2x}\implies$

$f(x)=\left(3\cot^2 x-1\right)-\left(1-3\tan^2x\right)=$ $3\left(\tan^2x+\frac 1{\tan^2x}\right)-2\ge 4$ $\implies$ $\boxed{\frac {\sin 3x}{\sin^3x}-\frac {\cos 3x}{\cos^3x}\ge 4}$ , with equality when $\tan^2x=1\iff x\in \pi\mathbb Z\pm\frac {\pi}{4}$ .

Proof 2. $\left\{\begin{array}{ccc}
\sin 3x=\sin x(2\cos 2x+1)\\\\
\cos 3x=\cos x(2\cos 2x-1)\end{array}\right\|\implies$ $E\equiv \frac {\sin 3x}{\sin^3x}-\frac {\cos 3x}{\cos^3x}=$ $\frac {2\cos 2x+1}{\sin^2x}-\frac {2\cos 2x-1}{\cos^2x}$ . Thus, $\left\{\begin{array}{ccc}
2\sin^2x=1-\cos 2x\\\\
2\cos^2x=1+\cos 2x\end{array}\right\|\implies$

$E=2\left(\frac {2\cos 2x+1}{1-\cos 2x}-\frac {2\cos 2x-1}{1+\cos 2x}\right)$ . Using the substitution $\cos 2x=t\in (-1,1)$ obtain that $E=2\left(\frac {2t+1}{1-t}-\frac {2t-1}{1+t}\right)=$ $\frac {4\left(2t^2+1\right)}{1-t^2}=$

$4\left(1+\frac {3t^2}{1-t^2}\right)\ge 4\implies$ $E\ge 4$ with equality if and only if $t=0$ , i.e. $\cos 2x=0$ .



PP8. Prove that for any real numbers $\{x,y\}$ there is the implication $\left\{\begin{array}{ccc}
x+y & = & m\\\\
\sqrt{x^2+1}+\sqrt{y^2+1} & = & n\end{array}\right|\implies$ $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\frac {n+m}{n-m}$.

Proof 1. Let$\left\{\begin{array}{c}
x+\sqrt{x^2+1}=\alpha\\\\
y+\sqrt{y^2+1}=\beta\end{array}\right|\implies$ $\left\{\begin{array}{c}
\alpha +\beta =m+n\\\\
\frac 1{\alpha}+\frac 1{\beta}=n-m\end{array}\right|$ $\implies$ $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\alpha\beta=$ $\frac {\alpha +\beta}{\frac 1{\alpha}+\frac 1{\beta}}=\frac {n+m}{n-m}\ .$

Proof 2 (trigonometric). Let $\left\{\begin{array}{ccc}
x=\tan a & \implies & \sqrt{x^2+1}=\frac 1{\cos a}\\\\
y=\tan y & \implies & \sqrt{y^2+1}=\frac 1{\cos b}\end{array}\right|$ . Thus, $\left\{\begin{array}{ccc}
x+y & = & m\\\\
\sqrt{x^2+1}+\sqrt{y^2+1} & = & n\end{array}\right|\implies$ $\left\{\begin{array}{ccc}
\tan a+\tan b & = & m\\\\
\frac 1{\cos a}+\frac 1{\cos b} & = & n\end{array}\right|\implies$

$\left\{\begin{array}{ccc}
\sin (a+b) & = & m\cos a\cos b\\\\
\cos a+\cos b & = & n\cos a\cos b\end{array}\right|\implies$ $\left\{\begin{array}{ccc}
2\sin\frac {a+b}2\cos\frac {a+b}{2} & = & m\cos a\cos b\\\\
2\cos\frac {a+b}2\cos\frac {a-b}2 & = & n\cos a\cos b\end{array}\right|\ (:)\ \implies$ $\frac {\sin\frac {a+b}2}{\cos\frac {a-b}2}=\frac mn\implies$ $\frac {\tan\frac a2+\tan\frac b2}{1+\tan\frac a2\tan\frac b2}=\frac mn\implies$

$\frac {n+m}{n-m}=\frac {\left(1+\tan\frac a2\right)\left(1+\tan\frac b2\right)}{\left(1-\tan\frac a2\right)\left(1-\tan\frac b2\right)}=$ $\frac {\left(1+\tan\frac a2\right)^2\left(1+\tan\frac b2\right)^2}{\left(1-\tan^2\frac a2\right)\left(1-\tan^2\frac b2\right)}=$ $\frac {\left(\cos\frac a2+\sin\frac a2\right)^2\left(\cos\frac a2+\sin\frac b2\right)^2}{\left(\cos^2\frac a2-\sin^2\frac a2\right)\left(\cos^2\frac b2-\sin^2\frac b2\right)}=$ $\frac {(1+\sin a)(1+\sin b)}{\cos a\cos b}=$ $\frac {1+\sin a}{\cos a}\cdot\frac {1+\sin b}{\cos b}=$

$\left(\tan a+\frac 1{\cos a}\right)\left(\tan b+\frac 1{\cos b}\right)\implies$ $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\frac {n+m}{n-m}$ .

Proof 3. Denote $\sqrt{x^2+1}-\sqrt {y^2+1}=mp$ . Thus, $\left\{\begin{array}{c}
\sqrt{x^2+1}+\sqrt {y^2+1}=n\\\\
\sqrt{x^2+1}-\sqrt {y^2+1}=mp\end{array}\right|\stackrel{(\pm )}{\implies}$ $\boxed{\begin{array}{c}
2\sqrt{x^2+1}=n+mp\\\\
2\sqrt{y^2+1}=n-mp\end{array}}\ (1)$ . Thus, $\left\{\begin{array}{c}
\sqrt{x^2+1}+\sqrt {y^2+1}=n\\\\
\sqrt{x^2+1}-\sqrt {y^2+1}=mp\end{array}\right|\bigodot\implies$

$\left\{\begin{array}{c}
x^2-y^2=mnp\\\\
x+y=m\end{array}\right|\implies$ $\left\{\begin{array}{c}
x+y=m\\\\
x-y=np\end{array}\right\|\implies$ $\boxed{\begin{array}{c}
2x=m+np\\\\
2y=m-np\end{array}}\ (2)$ . From the relations $(1)$ and $(2)$ obtain that $\left\{\begin{array}{c}
4\left(x^2+1\right)=(n+mp)^2\\\\
4x^2=(m+np)^2\end{array}\right|\implies$

$(m+np)^2+4=(n+mp)^2\implies$ $\left(n^2-m^2\right)p^2=n^2-m^2-4\implies$ $1-p^2=1-\frac {n^2-m^2-4}{n^2-m^2}\implies$ $\boxed{1-p^2=\frac 4{n^2-m^2}}\ (3)$ .

In conclusion, $4\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=$ $\left(2x+2\sqrt{x^2+1}\right)\left(2y+2\sqrt{y^2+1}\right)\ \stackrel{(1\wedge 2)}{=}$ $[(m+np)+(n+mp)][(m-np)+(n-mp)]=$

$(m+n)(1+p)\cdot (m+n)(1-p)\ \stackrel{(3)}{=}\ (m+n)^2$ $\cdot\frac 4{n^2-m^2}\implies$ $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\frac {n+m}{n-m}$



PP9. Prove that $\prod_{k=1}^{45}\left(1+\tan k^{\circ}\right)=2^{23}$ . Proof. Observe that $\left(1+\tan x\right)\left[1+\tan\left(45^{\circ}-x\right)\right]=2$ .


PP10. Solve the irrational equation $\frac {\sqrt {x+1}+\sqrt {x-1}}{\sqrt{x+1}-\sqrt {x-1}}=2x-\frac 12\ ,\ x\ge 1$ .

Proof. Let $\sqrt{\frac {x+1}{x-1}}=t\in (1,\infty )$ , i.e. $\frac {x+1}{t^2}=\frac {x-1}{1}=\frac {2x}{t^2+1}=\frac {2}{t^2-1}\implies$ $\boxed{x=\frac {t^2+1}{t^2-1}}\ (*)$ . Thus, $\frac {\sqrt {x+1}+\sqrt {x-1}}{\sqrt{x+1}-\sqrt {x-1}}=2x-\frac 12\iff$ $\frac {t+1}{t-1}=\frac {2\left(t^2+1\right)}{t^2-1}-\frac 12$

$\iff$ $\frac {t+1}{t-1}=\frac {3t^2+5}{2\left(t^2-1\right)}\iff$ $2(t+1)^2=3t^2+5\iff$ $t^2-4t+3=0\iff$ $\odot\begin{array}{ccccc}
\nearrow & t_1=1 & \not > & 1 & \searrow\\\\
\searrow & t_2=3 & > & 1 & \nearrow\end{array}\odot\iff $ $t=3\stackrel{(*)}{\iff}x=\frac {10}{8}\iff$ $\boxed{x=\frac 54}$ .



PP11. Find the real numbers $\{a,b,c\}$ so that $a^2+b^2+c^2=a^3+$ $b^3+c^3=a^4+b^4+c^4$ .

Proof. $\sum\left(a^4-2a^3+a^2\right)=0\iff$ $\sum \left(a^2-a\right)^2=0\iff$ $a^2-a=b^2-b=c^2-c=0\iff$ $\{a,b,c\}=\{0,1\}$ .


PP12. Fie $z\in\mathbb C^*$ for which $z+\frac 1z=i\sin\frac {\pi}{5}$ . Ascertain $z^{2014}+\frac 1{z^{2014}}$ . I think that the enunciation of this exercise is erroneously.

Proof. Observe that $z\ne 0$ . Thus, $\left\{\begin{array}{ccc}
z+\frac 1z & = & i\sin\frac {\pi}{5}\\\\
\overline z+\frac 1{\overline z} & = & -i\sin\frac {\pi}{5}\end{array}\right\|\bigoplus\implies (z+\overline z)+\left(\frac 1z+\frac 1{\overline z}\right)=0 \implies$ $(z+\overline z)\left(|z|^2+1\right)=0 \implies$ $z+\overline z=0\implies$

$\Re (z)=0\implies$ $(\exists )\ x\in\mathbb R^*$ so that $\boxed{z=ix}\implies$ $ix+\frac 1{ix}=i\sin\frac {\pi}{5}\implies$ $ix-\frac i{x}=i\sin\frac {\pi}{5}\implies$ $\boxed{x-\frac 1x=\sin\frac {\pi}{5}}$ , where $\boxed{\cos\frac {\pi}{5}=\frac {1+\sqrt 5}{4}\ \ \wedge\ \ \sin\frac {\pi}{10}=\frac {-1+\sqrt 5}{4}}$ .

Thus, $x^2+\frac 1{x^2}=\left(x-\frac 1x\right)^2+2=2+\sin^2\frac {\pi}{5}\implies$ $\boxed{y+\frac 1y=3-\cos^2\frac {\pi}{5}}=\frac {21-\sqrt 5}{8}$ , where $\boxed{y=x^2}$ . Must find $z^{2014}+\frac 1{z^{2014}}=-\left(y^{1007}+\frac 1{y^{1007}}\right)$ , what is easily a.s.o.



PP13. Polynomial $f
\in\mathbb Q[X]$ leaves remainder $15$ when divided by $(X-3)$ and remainder $(2X+1)$ when divided by $(X-1)^2$ . Find the remainder when divided by $(X-3)(X-1)^2$ .

Proof 1. The polynomial $f$ leaves remainder $(2X+1)$ when divided by $(X-1)^2\iff$ exists $g\in\mathbb Q[X]$ so that $\boxed{f=(X-1)^2\cdot g+(2X+1)}\ (*)$ . The polynomial $f$ leaves

remainder $15$ when divided by $(X-3)\iff f(3)=15\ \stackrel{(*)}{\iff}$ $4g(3)+7=15\iff$ $\boxed{g(3)=2}\iff$ the polynomial $g$ leaves remainder $2$ when divided by $(X-3)\iff$

exists $h\in\mathbb Q[X]$ so that $\boxed{g=(X-3)\cdot h+2}$ . In conclusion, the required polynomials $f$ with given properties are $f\ \stackrel{(*)}{=}\ (X-1)^2\cdot \left[(X-3)\cdot h+2\right]+(2X+1)=$

$(X-3)(X-1)^2\cdot h+2(X-1)^2+2X+1\iff$ $\boxed{f=(X-3)(X-1)^2\cdot h+\left(2X^2-2X+3\right)}$ , where $h\in\mathbb Q[X]$ .

Proof 2. I"ll seek the remainder $r=aX^2+bX+c$ of the division $f$ by $(x-3)(x-1)^2$ , i.e. $f=(X-3)(X-1)^2\cdot h+r$ . Therefore,

$\left\{\begin{array}{ccccc}
f(3)=15 & \implies & r(3)=15 & \implies & 9a+3b+c=15\\\\
f(1)=3 & \implies & r(1)=3 & \implies & a+b+c=3\\\\
f'(1)=2 & \implies & r'(1)=2 & \implies & 2a+b=2\end{array}\right\|\implies$ $\left|\begin{array}{c}
a=2\\\\
b=-2\\\\
c=3\end{array}\right|\implies$ $\left\{\begin{array}{cccc}
r=2X^2-2X+3\ ;\ h\in\mathbb Q[X]\\\\
f=(X-3)(X-1)^2\cdot h+r\end{array}\right\|$.



PP14. Let the "triangle" of all odd natural numbers $\left\{\begin{array}{ccc}
L_1 & : &  1\\\
L_2 & : & 3\ \ 5\\\
L_3 & : & 7\ \ 9\ 11\\\
L_4 & : & 13\ 15\ 17\ 19\\\
\ldots & : & \ldots\ \ldots\ \ldots\ \ldots\end{array}\right\|$ . Ascertain $:$

$\blacktriangleright$ The first term and the last term of the $\mathrm{n^{th}}$ line, where $n\in \mathbb N^*\ ;$

$\blacktriangleright$ The sum of the terms from the $\mathrm{n^{th}}$ line $\ ;$

$\blacktriangleright$ Prove that $1^3+2^3+\ \ldots\ +(n-1)^3+n^3=\frac {n^2(n+1)^2}4=\left[1+2+3+\ \ldots\ +(n-1)+n\right]^2\ ;$

$\blacktriangleright$ Find the "coordinates" of the number $1001$ , i.e. find the line that belongs to and where is number $1001$ in this line.


Proof. Denote the sequence of all odd natural numbers $a_p=2p-1$ , where $p\in\mathbb N^*$ and the terms $b_{nk}$ of the line $L_n$ , where $k\in\overline{1,n}$ .

$\blacktriangleright$ Observe that $:$ the line $L_n$ has $n$ terms $;$ its first term is $a_{1+2+\ \ldots\ +(n-1)+1}=$ $a_{\frac {n(n-1)}2+1}=$ $2\left[\frac {n(n-1)}2+1\right]-1=$ $n^2-n+1\implies $

$\boxed{b_{n1}=n^2-n+1}\ ;$ its last term is $a_{1+2+\ \ldots\ +(n-1)+n}=$ $a_{\frac {n(n+1)}2}=$ $2\left[\frac {n(n+1)}2\right]-1=$ $n^2+n-1\implies \boxed{b_{nn}=n^2+n-1}\ .$

$\blacktriangleright$ The sum $S_n$ of the terms from the $\mathrm{n^{th}}$ line $L_n$ is $\sum_{k=1}^n b_{nk}=\frac {b_{n1}+b_{nn}}2\cdot n=\frac {2n^2}2\cdot n\implies$ $\boxed{S_n=n^3}$ for any $n\in\mathbb N^*$ .

In conclusion, $\sum_{k=1}^nk^3=\sum_{k=1}^nS_k=$ $\sum_{p=1}^{1+2+\ \ldots\ +n}(2p-1)=(1+2+\ \ldots\ +n)^2$ because is well-known that $\sum_{k=1}^n(2k-1)=n^2$ .

$\blacktriangleright n^2-n+1\le 1001\le n^2+n-1\iff$ $n^2-n+1\le 1001< n^2+n\iff$ $n=32$ and $L_{32}\ :\ 993\ ,\ 995\ ,\ 997\ ,\ 999\ ,\ 1001\ \ldots\ 1053\ ,\ 1055$ , i.e. $\boxed{1001=b_{32;5}}$ .

Remark. For an odd $p\in\mathbb N^*$ seek $n\in\mathbb N^*$ so that $p\in L_n$ and the position $k$ of $p\equiv b_{nk}$ in the row $L_n$ . Therefore, $n^2-n+1\le p\le n^2+n-1\iff$ $n^2-n<p<n^2+n\iff$

$\frac {-1+\sqrt{4p+1}}2<n<\frac {1+\sqrt{4p+1}}2\iff$ $n-1\le \frac {-1+\sqrt{4p+1}}2<n\iff$ $n-1=\left[\frac {-1+\sqrt{4p+1}}2\right]\iff$ $n=1+\left[\frac {-1+\sqrt{4p+1}}2\right]$ . Observe that in the row

$L_n\ ,\ n\in\mathbb N^*$ have $p=\left(n^2-n+1\right)+2(k-1)\implies$ $\boxed{k=\frac {p-n^2+n+1}2}$ ... Otherwise. In the row $a_n\ ,\ n\in\mathbb N^*$ have $p=a_{k+\frac {n(n-1)}2}=1+2\left[\frac {n(n-1)}2+k-1\right]=$

$1+n(n-1)+2k-2\implies$ $p=2k+n(n-1)-1\implies$ $k=\frac {p-n^2+n+1}2\ .$
This post has been edited 171 times. Last edited by Virgil Nicula, Feb 11, 2018, 3:04 PM

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    by ryanbear, May 9, 2023, 6:10 AM

  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

    by tigerzhang, Aug 1, 2021, 12:02 AM

  • Amazing blog.

    by OlympusHero, May 13, 2021, 10:23 PM

  • the visits tho

    by GoogleNebula, Apr 14, 2021, 5:25 AM

  • Bro this blog is ripped

    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404398
  • Total comments: 37
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