2011/067) factor $a^6 + 18a^3 + 125$

the above is not difficult though it looks so

we realize $a^6 = (a^2)^3$ and $125 = 5^3$

now if we can split (is it possible) 18 so that $18 = x^3 + y$ and  $-y = 15 x$  (which is true x = 3 and y = -45)

we get $a^6 + 18a^3 + 125 = a^6 + 27 a^3 - 45 a^3 + 125$
$= (a^2)^3 + (3a)^2 + 5^3 - 3(a^2)(3a) 5$

and using $x^3 + y^3 + z^3 = (x+y+z)(x^2+y^2+z^2- xy-yz-zx)$
we get
$(a² - 3a + 5)((a^4) + 9a^2 + 25 + 3(a^3) + 15a - 5a²)$

$= (a² - 3a + 5)((a^4) + 3(a^3) + 4a² + 15a + 25)$

2011/066) find near issoceles triangles

that is the 2 legs ar of length x, x+ 1
let legs be x and x+1 and hypotenuse y

x^2+ (x+1)^2 = y^2
or 2x^2 + 2x + 1 = y^2
or doubling

(4x^2+4x+2) = 2y^2
or (2x+1)^2 = 2y^2 -1
putting 2x +1 = m
we get 2y^2-m ^2= 1

now by trial we see that y = 5 and m = 7 is a solution
if (a,b) is a solution then (2a^2-b^2) = 1

so (a sqrt(2) – b) (a sqrt(2) + b) = 1
so | (a sqrt(2) – b) | = 1
cubing both sides ( sqaring shall not help as we get ( a+b- 2ab(sqrt(2)) this arises as |1| = |-1| = 1
(2a^3 + 3ab^2) sqrt(2) –(12a^2 + b^3) (the sqrt(2) term is positive and the non sqrt(2) is – ve)
So it is a solution
Starting with (5,7) we get (985,1393) or the triple is ( 696,697,845)
So we can get the terms as far as possible.
But It shall not give all solutions

2011/065) If sides of length a b c form a triangle prove sqrt(a), sqrt(b), sqrt(c) form a triangle

Let's assume a >= b >= c > 0. As a, b, c are sides of a triangle, they fit the triangle inequality a < b + c.

(1) (sqrt(b) + sqrt(c))^2 = b + 2*sqrt(b)*sqrt(c) + c

Value of sqrt(x) is positive for any positive x, so

(2) (sqrt(b) + sqrt(c))^2 > b + c > a

As square root is an increasing function, (2) follows

(3) sqrt(b) + sqrt(c) > sqrt(a)

Hence proved