2021/047) Find all natural numbers n such that $n^2+7| n^3 + 3$

We have $n^3+3 = n(n^2+7) - (7n-3)$

Because $n^2+7| n^3 + 3$ so $n^2+ 7 | 7n - 3$

As 7n - 3 is positive so we have

$n^2 + 7 \le 7n- 3$

Or $n^2 - 7n + 4 \le 0$ giving $n < 7$

By trying the values of n from 1 to 6 we get  1

Solution set $\{2,5\}$ 

2021/046) Find all 2 digit numbers having 6 factors

we shall be using the following formula for number of factors of a number.

If the number N is of the form $p_1^{q_1}p_2^{q_2} \cdots p_n^{q_n}$ when 

$p_1,p_2,p_3\cdots p_n$ are prime numbers and

 $q_1,q_2\cdots q_n$ are the powers of the prime numbers then the number of factors 

 are $(q_1+1)(q_2+1)\cdots(q_n+1)$  

now let us factor 6 is different ways 6 = 6 = 2 * 3

if we take 6 then the number is $p^5$ and for this to be between 10 and 100 this is $2^5= 32$ as $3^5 = 243$ is bigger

If we take 2* 3 then the number is $p_1p_2^2$ and for this to be between 10 and 100 

Taking $p_1=2$ we get $2* 3^2= 18$, $2 * 5^2 = 50$, $2 * 7^2 = 98$.

Taking $p_1=3$ we get $3* 2^2= 12$, $3 * 5^2 = 75$

Taking $p_1=5$ we get $5* 2^2= 20$, $5 * 3^2 = 45$

Taking $p_1=7$ we get $7* 2^2= 28$, $7 * 3^2 = 63$

Taking $p_1=11$ we get $11* 2^2= 44$, $11 * 3^2 = 99$

Taking $p_1= 13$ we get $13* 2^2= 52$

Taking $p_1= 17$ we get $17* 2^2= 68$

Taking $p_1= 19$ we get $19* 2^2= 76$

Taking $p_1= 23$ we get $23* 2^2= 92$

So solution set 

$\{12,18,20,28, 32,44,45,50,52,63,68,75,76,92,98,99\}$

2021/045) Let $p>5$ be the prime number. Prove that the expression $p^4-10p^2+9$ is divisible by 1920?

 we need to show that $p^4 - 10p^2 +9)$ is divisible by 1920.

Let us now factor $p^4 - 10p^2 +9$

$p^4 - 10p^2 +9 = (p^2-1)(p^2-9)$

$= (p+1)(p-1)(p+3)(p-3)$

$= (p-3)(p-1)(p+1)(p+3)$

as p is greater than 5 and a prime so p is odd so let p = 2n + 1

so we get above expression

$=(2n-2)(2n)(2n+2)(2n+4) = 16(n-1)n(n+1)(n+2)$

$(n-1)n(n+1)(n+2)$ being product of 4 consecutive numbers is divisible by 24 so $16(n-1)n(n+1)(n+2)$ is divisible by 16 * 24 = 384

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

$(n-1)n(n+1)(n+2)(n+3)$ being product of 5 consecutive numbers is divisible by 5 but is p is prime so p is not

 divisible by 5 or 2n+1 is not divisible by 5 or 2n +6 is not divisible 5 or n +3 is not divisible by 5 so $(n-1)n(n+1)(n+2)$ is not divisible by 5

Hence $16(n-1)n(n+1)(n+2)$ is divisible by 5 and is it is divisible by 384 so divisible by 5 * 384 or 1920

so $p^4 - 10p^2 +9$ is divisible by 1920 hence proved.

2021/044) Prove that $13^{99} - 19^{93}$ is divisible by 162

we have $162= 2 * 81 =2 * 3^4$

So we need to show that the given expression is divisible by 2 and 81

Divisible by 2 is simple and both terms are odd so the difference is even so divisible by 2

Now let us find the value of expression mod 81

To find the same let us evaluate each term mod 81

$13^{99}= (12+1)^{99} = \sum_{n=0}^{99}{99 \choose n}12^n 1^{99-n}$

for n = 4 to 99 there is power of $12^n$ so each of the terms is divisible by $3^4$ or 81

so we have $13^{99} \equiv 1 + 99 * 12 + \frac{99 *98}{2} * 12^2 +   \frac{99 *98 * 96 }{6} * 12^3 \pmod {81}cdots(1)$

out of above the 3rd term onwards divisible by 81( as it contains 99 so one 9 is there so we require another 9 that comes from $12^2$ so taking mod 81 all the terms from 3rd term onwards is zero

So

$13^{99} \equiv 1 + 99 * 12 =  1+ (81 * 18) * 12 \equiv 1 + 18 * 12 \equiv 217 \equiv 55 \pmod {81}$

Now

$19^{93}= (18+1)^{93} = \sum_{n=0}^{93}{93 \choose n}18^n 1^{93-n}= 1 + 18 * 92 + \sum_{n=2}^{93}{93 \choose n}18^n 1^{93-n}$

except 1st 2 terms all are divisible by 81 so we have

$19^{93}\equiv 1 + 18 * 93 \equiv 1 + 18 * 12 \equiv 217 \equiv 55  \pmod {81}\cdots(2)$

from (1) and (2)

$13^{99} - 19 ^{93} \equiv 0 \pmod {81}$

as $13^{99} - 19 ^{93} \equiv 0 \pmod {2}$

from above 2 we get $13^{99} - 19 ^{93} \equiv 0 \pmod {162}$  

2021/043) Find the largest integer n for which $\sum_{k=1}^\infty \frac{1}{k^n}$ diverges

Ans n = 1.

To prove that same we show that if n = 1 then this diverges and if n =2 this converges

To prove that it diverges for n = 1

Let the sum be S

We have $S = \sum_{k=1}^\infty \frac{1}{k^n}$

Or 

$S = 1 +  \sum_{k=1}^\infty (\sum_{i=2^{k-1}+1}^{2^{k}} \frac{1}{i^n})$

 $S = 1 + \sum_{k=1}^\infty s_k\cdots(1)$

where $s_k = \sum_{i=2^{k-1} +1}^{2^{k}} \frac{1}{i^n}$

Basically we have broken the sum to smaller groups with each group $k^{th}$ group containing the sum of reciprocal of $2^{k-1} +1$ to $2^{k}$ we have separated 1 and 1/2 so that k can start from 1 

Now let us compute $s_k = \sum_{i=2^{k-1} +1}^{2^{k}} \frac{1}{i^n}$

The above has got $2^{k-1}$ terms 

and for $i < 2^k$ we have $\frac{1}{i} > \frac{ 1}{2^k}$

Hence $s_k >= 2^{k-1} * \frac{ 1}{2^k}$

or $s_k >= \frac{ 1}{2}$

Putting in (1) we get

$S = 1 + \sum_{k=1}^\infty s_k = 1 + \sum_{k=1}^\infty \frac{1}{2}$

Hence it diverges 

Now let us prove that it converges for n =2

For $n >=2$

We have $n^2 > n(n-1)$ or $\frac{1}{n^2} < \frac{1}{n(n-1)}$

We have $n^2 > n(n-1)$ or $\frac{1}{n^2} < \frac{1}{n(n-1)}$

as $\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$

so $\frac{1}{n^2} <  \frac{1}{n-1} - \frac{1}{n}$

So $\sum_{k=1}^\infty \frac{1}{k^n}$

$= 1 + \sum_{k=2}^\infty \frac{1}{k^n}$

$<  1 + \sum_{k=2}^\infty(\frac{1}{k-1} - \frac{1}{k}$

$<= 1 + (1- \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3} ) + \cdots$

$< 2$

 

2021/042) For real numbers a and b that satisfy $a^3+12a^2+49a + 69=0$ and $b^3 -9b^2+ 28b -31 = 0$ find a+b.

We have
 $a^3+12a^2+49a + 69=0$
=> $(a+4)^3 + (a+5) = 0$
or $(a+4)^3 + (a+4) + 1= 0\cdots(1)$

$b^3 -9b^2+ 28b -31 = 0$
$=>(b-3)^3 + (b-4) = 0$
$=>(b-3)^3 + (b-3) -1 = 0\cdots(2)$

if we define $f(x) = x^3+x$ knowing that f(x) is odd function
$(f(a+4) = -1$ and $-f(b-3) = 1$

so $(a+4) = - (b- 3)$

or $a + b = - 1$

2021/041) The equation $(x+a)(x+b)=9$ has a root a+b . Prove that $ab \le 1$

 Solution 

Because a+b is a root so $(x+a)(x+b)=9$ shall be satisfied when x = a + b.
Putting x = a + b we get
$(2a+b)(2b+a) = 9$
or $2a^2 + 2b^ 2 + 5ab = 9$
or $2(a^2+b^2) + 5ab = 9$
or $2(a^2+b^2-2ab) + 9ab = 9$
or $2(a- b)^2 = 9(1-ab)$
now LHS is >=0 so is RHS so $9(1-ab) \ge 0$ or $1-ab \ge 0$ or $ab \le 1$

2021/040) Prove that there are infinitely many positive integers n such that n(n+1) can be expressed as a sum of two squares in atleast two different ways.

 Let the number be $z = n(n+1)$

So $z = n^2 + n$

If we choose n to be a square say $m^2$ then we have z already a sum of 2 squares

We have $z = (m^2)^2 + m^2= m^2(m^2+1)$

If we have $m^2$ as sum of 2 squares say $x^2+y^2$ then 

we have

$z = (p^2 + q^2) (m^2 + 1) = p^2m^2 + q^2m^2 + p^2 + q^2$

$= (p^2m^2 + q^2 - 2pqm) + (q^2m^2 +p^2 + 2pqm)$

$= (pm - q)^2 + (qm +p)^2$

This is another way of representation

As (p,q,m) are sides of a Pythagorean triple and there are infinite Pythagorean triples so there are infinite independent values.

2021/039) if $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ where a,b,c are positive integers with no common factor then prove that a+b is a perfect square

 We have 

$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$

or $bc + ac = ab$

or $(a-c)(b-c)= c^2$

Now let a prime divide c 

There are two cases 

p divides a-c and b-c in which case p divides a,b,c so they have a common factor p which cannot be true

or $p^2$ divides a-c or b-c so there exists m and n(either of them can be 1) such that

$c= mn$ and $(a-c) = m^2$ and $(b-c) = n^2$   

so $a = m^2 + c$  and $b = n^2 +c$

so $a+b= m^2 + n^2 + 2c = m^2 + n^2 + 2mn = (m+n)^2$

or $a+b$ is a perfect square