2017/020) Find n such that both $n+3$ and $n^2+3$ are perfect cubes

If $(n+3)$  and $n^2+3$ are perfect cubes then their product is a perfect cube say $x^3$

$x^3 = (n+3)(n^2+3) = n^3 + 3n^3 + 3n + 9 = (n+1)^3 + 8 = (n+1)^3 + 2^3$

$x^3 = (n+1)^3 + 2^3$ this does not have a non trivial solution

so x = 0 or n = -1 and n= -1 => n+3 is 2 which is not a cube
x = 0 => n+1 = - 2 so $n^2  + 3 =  12$ which is not a cube

so no solution

2017/019) Rational or Irrational

Let a_n be defined as follows for all natural numbers n:

a_n = 0 if the number of divisors of n (including 1 and n) is odd
a_n = 1 otherwise.

Now consider the fraction 0.a₁a₂a₃....

Is this fraction rational or irrational? Explain.

Solution:

the number of factors is odd for square number and it is even for non square numbers.

so $a_n = 0$ for no square number and = 1 for no square number

so in the decimal at each square place it is 1 and the gaps keeps on increasing.

so the digits do not recur and hence it is irrational

2017/018) ABC is a triangle and O is a point in it. Prove that $(AB + BC + AC ) > (OA + OB + OC)$

Let BO meet AC in D

Then $AB+AD > BD = OB+OD$

And $ OD+DC > OC$

Sum these for $AB+(AD+DC)+OD > OC+OB+OD$

or $AB+AC > OC+OB$

Similarly for BA+BC and CB+CA and sum to get

$2(AB+BC+CA) > 2(OA+OB+OC)$

or $AB+BC+CA > OA+OB+OC$

2017/017) Find positive n such that $lim_{x->3} \frac{x^n-3^n}{x-3} = 108$

we have from definition LHS = $\frac{dy}{dx}$ at x = 3 where $y=x^n$
so differentiating we get $(n)3^{n-1} = 108 = 4 * 3^3$ or n = 4

2017/016) Find a,b satisfying $4^a + 4a^2 + 4 = b^2$

The LHS is even hence to put a limit on a we must have
$(2^a+2)^2 <= 4^a + 4a^2 + 4$
or $4^a + 4 * 2^a + 4 <= 4^a + 4a^2 + 4$ or $2^a <= a^2$ or $ a <= 4$
putting a = 0 to 4 one by one we get $a=2,b=6$ and $a=4,b= 18$ are 2 solution

2017/015) Show that there are infinitely many n such that $6n+1$ and $6n-1$ are both are composites

we have for n > 2 $n^3-1= (n-1)(n^2+n+1)$ is composite and $n^3+1=(n+1)(n^2-n+1)$ is composite for n ,
so if we  choose $n= 36m^3$ we get $6n-1= (6m)^3-1$ and $6n+1= (6m)^3 + 1$ composites