2017/003) Show that for a,b,c all different $\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a} != 0$

if  $\sqrt[3]{a-b} + \sqrt[3]{b-c } + \sqrt[3]{c-a} = 0$
then using $(x + y + z = 0) =>x^3 + y^3 + z^3 = 3xyz$
we get $0 = 3\sqrt[3]{(a-b)(b-c)(c-a)}$ giving a = b or b= c or c=a.
so if all are different result cannot be zero.

2017/002) Given a cubic eqaution $x^3 + px^2+ qx+r =0$ has 2 roots

 that are equal in value and opposite in sign show that $pq=r$

Proof:

Let the roots be $a,-a,b$
so we have $x^3+px^2+qx + r = (x-a)(x + a)(x-b) = (x^2 - a^2)(x-b) = x^3 - b x^2 -a^2 x + ba^2$
giving $p = -b$, $q=-a^2$, $r = ba^2$
hence $pq = a^2b = r$

2017/001) Find all values of the parameter a such that the real roots $\alpha,\beta$ of the equation $2x^2+6x+a=0$ satisfy the ineqauilty $|\frac{\alpha}{\beta} + \frac{\beta}{\alpha} |< 2$

as  for real t $|t+\frac{1}{t}| >=2$
so one root is positive and another -ve
$2x^2+6x + a = 0$
$=>4x^2 + 12x + 2a = 0$
$=>(2x+3)^2 + 2a - 9 = 0$
$=>(2x+3)^2 =  9 - 2a$
$x = -\frac{3}{2} \pm \frac{\sqrt{9-2a}}{2}$
for one root to be positive and another -ve $\frac{\sqrt{9-2a}}{2} > \frac{3}{2}$ or $ a < 0$
if we were interested in complex roots then $a > \frac{9}{2}$

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.