2021/111) For which primes p, 7p+4 is a perfect square?

 7p + 4 is a perfect square say $m^2$

so $7p = m^2 -4 = (m+2)(m-2)$

now there are 2 cases

1. p is 2 which gives 7p + 4 = 18 which is not a perfect square

so

2. p is odd

so 7p is odd and  $m^2 -4$ is odd

so m+2 and m-2 are co-primes as they differ by 4

so m+2 = 7, m-2 = p gives p = 3 which is prime and m = 5

or m+2 = p and m-2 = 7 giving m = 9 and p =11 which is a prime

so p = 3 or 11 

2021/110) express $\cos\, 5t$ in term of power of $\cos\, t$

We have

$\cos\, 5t = (\cos\, 5 t + \cos\, t) - \cos\, t$

$= 2 \cos\, 3t \cos\, 2t - \cos\, t$ using $\cos\, A + \cos\, B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

$= 2 * (4 \cos ^3 t - 3 \cos\, t)(2\cos ^2 t - 1) - \cos\, t$ using formula for $cos 3t$ and $cos 2t$

$= 16 \cos^5 t - 20 \cos^3 t + 6 \cos t -\cos t$

$= 16 \cos^5 t - 20 \cos^3 t + 5 \cos t$

2021/109) Find the sum of the series $\sum_{n=1}^{\infty}\frac{1}{n^2}$

We have tailor  expansion of $\sin\, x$  as  

$P(x) = \sin\,x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots  (1) $

This is a polynomial of degree infinite with zeroes at  0 and npi so

this is  $Ax(1- \frac{x}{\pi})(1+ \frac{x}{pi})(1-\frac{x}{2pi})(1-\frac{x}{2\pi})\cdots$

or $P(x) =Ax(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

comparing above with (1) we get A = 1

So $P(x) =x(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

The coefficient of $x^3$ is    $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2}$

from (1)  coefficient of $x^3$ is $-\frac{1}{6}$

as both are same so  $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2} = -\frac{1}{6}$

multiplying both sdes by $- \pi^2$ we get

 $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ 

2021/108) The polynomial: $P(x) = 1 + a_1x +a_2x^2+...+a_{n-1}x^{n-1}+x^n$ with non-negative integer coefficients has $n$ real roots. Prove, that $P(2) \ge 3^{n}$

Because all coefficients are positive so all n roots are -ve and hence

$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive

Further $\prod_{k=1}^{n} (a_k) = 1$

So $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$

Now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$

So from (1) and (2)

$P(2) >= 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)}  = 3^n$ and hence $P(2) >= 3^n$

2021/107) Show that $\tan^{-1}(k) = \sum_{n=0}^{k-1}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right )$ - and deduce that $\sum_{n=0}^{\infty}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ) = \frac{\pi}{2}$

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$

Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$

Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$

We get  $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)-  \tan ^{-1}n$

Adding from 0 to k-1 we get as telescopic sum

Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} =  \tan ^{-1}k-  \tan ^{-1}0 = \tan ^{-1}k$

Taking limit as $k = \infty$

$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1}  = \tan ^{-1}\infty= \frac{\pi}{2}$

2021/106) Evaluate closed form of $1^2+2^2+3^2+\cdots+n^2$

We have

$(k+1)^3 = k^ 3 + 3k ^2 + 3k + 1$

Or $(k+1)^3 - k^ 3 = 3k ^2 + 3k + 1$

Adding from 1 to n we get

$\sum_{k=1}^n((k+1)^3 - k^ 3) = 3\sum_{k=1}^n k^2 + 3\sum_{k=1}^nk + \sum_{k=1}^n1$

The LHS is a telespcopic sum = $(n+1)^3-1$

We know $\sum_{k=1}^nk = \frac{n(n+1)}{2}$

so we get $(n+1)^3 - 1 = 3\sum_{k=1}^n k^2 + 3\frac{n(n+1)}{2} + n$

or  $3\sum_{k=1}^n k^2 = (n+1)^3 - 1 -  3\frac{n(n+1)}{2} - n$

$=n^3 + 3n^2 + 3n + 1 - 3\frac{n^2+n}{2} - n$

$= \frac{1}{2}(2n^3 + 6n^2 + 6n - 3n^2 -3n -n )$

$= \frac{1}{2}(2n^3 + 3n^2 + 2n )$

$=\frac{1}{2}n(2n^2+ 3n + 2)$

$= \frac{1}{2}n(n+2)(2n+1)$

so $\sum_{k=1}^n k^2 = \frac{1}{6}(n(n+2)(2n+1)$

2021/105) Prove that if k = mn and k is a perfect square and m and n are co-primes them m and n are perfect squares

 Now let p be a prime factor of k.

So p is a prime factor of m or n but not both because GCD(m, n) = 1

Now because k is a square p shall occur even number of times say 2m

All the 2m occurences must be factor of m (as we have mention p is factor of m) 

So any prime factor of k whcich is a factor of m shall occur even number of times in m and which is not a factor of m shall occur even number of times in n making both m and n perfect squares.