2016/121) Express $(1+a^2)(1+b^2)$ as sum of 2 squares

we have $(1+a^2)(1+b^2) = (1+ ia)(1-ia)(1+ib)(1-ib)$ factoring over complex
$=((1+ia)(1+ib))(1-ia)(1-ib))$
$=(( 1 - ab) + i(a+b))((1 - ab) - i(a+b)) = (1- ab)^2 + (a+b)^2$ multiplication over complex

2016/120) Solve in positive integers $p^x = y^4 + 4$ and p is prime

factor the RHS ro get $p^x = (y^2-2y+2)(y^2+2y+2)$
both are power of p and $y^2-2y+2)< y^2 + 2y + 2$ so $y^2-2y + 2$ is a factor of difference that is 4y
so $y^2-2y+2$ is a factor of $4y^2$ or $4y^2-4(y^2-2y+2) = 8(y-1)$
so $y^2-2y+2$ is a factor of 8 that is it is 1 or 2 or 4 or 8
if it is 1 then y = 1, p = 5 and x = 1
if it is 2 then y=2 giving $p^x = 20$ which is not possible and if it is 4 or 8 we do not have integer solution
so only solution is $p = 5, x = y= 1$

2016/119)show that if A, B,C are angles of triangle $\tan\, A \tan\, B \tan\, C>=3\sqrt3$

because A B and C are angles of a triangle we have
$A+B= (180^\circ-C)$
taking tan of both sides
$\tan (A + B) = -\tan \, C$
or $\frac{\tan\, A + \tan \, B}{1- \tan\, A \tan\, B} = -\tan \, C$
or $\tan \, A + \tan \, B = - \tan C + \tan\, A \tan\, B \tan \, C$
or  $\tan \, A + \tan \, B + \tan C =  \tan\, A \tan\, B \tan \, C$
so $\tan\, A \tan\, B \tan \, C = \tan \, A + \tan \, B + \tan C$
using AM GM inequalty we have
$\frac{\tan \, A + \tan \, B + \tan C}{3} >= \sqrt[3]{\tan \, A  \tan \, B + \tan C}$
or  $\frac{\tan \, A  \tan \, B  \tan C}{3} >= \sqrt[3]{\tan \, A  \tan \, B + \tan C}$
or  $\tan \, A  \tan \, B  \tan C >= 3 \sqrt[3]{\tan \, A  \tan \, B  \tan\, C}$
cube both sides to get $(\tan \, A  \tan \, B  \tan C)^3 >= 27 (\tan \, A  \tan \, B + \tan C)$
or $(\tan \, A  \tan \, B  \tan C)^2 >= 27 $
or $(\tan \, A  \tan \, B  \tan C) >= 3\sqrt{3}$

2016/118) Let X be set of +ve integers greater than or equal to 8 and f(x) X->X be a function such that f(x+y) = f(x)f(y).

we have f(9) = f( 4 + 5) = f(4 * 5) = f(20) = f( 4+ 16) = f(64) = f( 8 + 8) = f(16) = f( 4 * 4) = f(8) = 8
in the above each step is valid because x,y are >=4.

2016/117) If a,b,c are 3 positive numbers show that $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} \frac{c^2+1}{a+b} >= 3$

we have $a^2+1 >= 2a$ hence $\frac{a^2+1}{b+c} >= 2\frac{a}{b+c}$
also $b^2+1 >= 2b$ hence $\frac{b^2+1}{c+a} >= 2\frac{b}{b+c}$
and $c^2+1 >= 2c$ hence $\frac{c^2+1}{a+b} >= 2\frac{c}{a+b}$
adding above 3 we get $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} \frac{c^2+1}{a+b} >= 2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})$
now
If we can show that $2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})>=3$
we are through
we have $\frac{a}{b+c} + \frac{b}{c+a}$
= $\frac{ca + a^2 + b^2 + bc}{c+a}$
$>=\frac{ca + 2ab + b^2 + bc}{c+a}$ since $a^2+b^2>=2ab$
$= \frac{a(b+c) + b(c+a)}{(c+a)}$
Thus  $\frac{a}{b+c} + \frac{b}{c+a} >= \frac{a}{a+c} + frac{b}{b+c}$
Similarly, it can be proved that
$\frac{b}{c+a} + \frac{c}{a+b} >= \frac{b}{a+b} + frac{c}{c+a}$
and $\frac{c}{a+b} + \frac{a}{b+c} >= \frac{c}{b+c} + frac{a}{a+b}$
Adding corresponding sides of the above three inequalities, we get
$2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})>=3$

2016/116) Find the real roots of the equation

$\sqrt{x+3-4\sqrt{x-1}} + \sqrt{x+8-6\sqrt{x-1}} = 1$


Solution


let $x-1 = y^2$
so we get  $\sqrt{y^2+4-4y} + \sqrt{y^2+9-6y} = 1$
or $\sqrt{(y-2)^2} + \sqrt{(y-3)^2} = 1$
or $|y-2| + |y-3| = 1$
as 3-2 is 1 so $2 <=y <=3$ or $5  <= x <= 10$

2016/115) For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove x,y.z can be sides of a triangle.

letting $x=\tan\,A$, $y=\tan\, B$, $z\tan\, C$ we have A, B, C, between 0 and $\frac{\pi}{4}$
also $\tan\, A \tan\,  B +  \tan\, B \tan\,  C + \tan\, C \tan\,  A = 1$
using $\tan(A+B+C)$ we get
$A+B+C=\frac{\pi}{2}$
so $B+C>=A$ from above and A between 0 and $\frac{\pi}{4}$
so $x <= y+z$
similarly $y <= z + x$ and $z <= x + y$ so x,y,z can be sides of a triangle