2021/068) Prove that the polynomial equation $x^8-x^7+x^2-x+15=0$ has no real solution.

 We have $x^8-x^7 + x^2 -x + 15 = x^7(x-1) + x(x-1) + 15$

Each term is positive for $x > 1$ so LHS is greater than 0 so no solution for $ x > 1$

For x = 1 LHS = 15 so x = 1 is not a solution

Further $x^8-x^7 + x^2 -x + 15 = (15- x) + x^2(1-x^5) + x^8 $

Each term is positive for $x < 1$ so LHS is greater than 0 so no solution for $ x < 1$

Hence no real solution

2021/067) Let a,b,c be three distinct integers and P be a polynomial with integer coefficients. Show that in this case the conditions P(a)=b,P(b)=c,P(c)=a cannot be satisfied simultaneously.

The polynomial is P(x)

we have m-n divides  P(m) - p(n)

Let the given condition is true

So $a-b | P(a) - p(b)$

or $a-b | b- c$

Similarly

$b - c | c- a$

And $ c- a | a - b$

From above 3 have

$a-b | b-c | c- a | a-b$

So all are same hence a = b= c which is a contradiction.

Or they are -1/+1 and this is also a contradiction 

so condition can not be satisfied simultaneously 

2021/066) Prove that two triangles with sides $a,\,b,\,c$ and $a_1,\,b_1,\,c_1$ are similar if and only if $\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a+b+c)(a_1+b_1+c_1)}$.

we have

$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a+b+c)(a_1+b_1+c_1)}$.

$\equiv (\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1})^2=(a+b+c)(a_1+b_1+c_1)$

$\equiv aa_1+bb_1+cc_1+2\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1} =  aa_1+ab_1 + ac_1 + ba_1 + bb_1 + bc_1 + ca_1 + cb_1 + cc_1$

$\equiv 2\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1} = ab_1 + ac_1 + ba_1 + bc_1 + ca_1 + cb_1$

$\equiv  ab_1 + ac_1 + ba_1 + bc_1 + ca_1 + cb_1-2(\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1}) = 0$

$\equiv  (\sqrt{ab_1} - \sqrt{a_1b})^2 +  (\sqrt{ac_1} - \sqrt{a_1c})^2 +  (\sqrt{bc_1} - \sqrt{b_1c})^2 = 0$

The above is true iff $ab_1 = a_1b$, $ac_1 = a_1c$, $bc_1 = b_1c$

giving $\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1}$ or the 2 triangles are similar

2021/065) Solve the biquadratic equation $(x-9)(x-7)(x+3)(x+5)= 1792$

Because (-7) -(-9) = 2 and and 5-3 = 3 let us take y to be mean of x- 7 and x + 3 that is x - 2

So get get

$(y-7)(y-5)(y+5)(y+7) = 1792$

or $(y-7)(y+7)(y-5)(y+5) = 1792$

or $(y^2-49)(y^2-25) = 1792$

or $y^4 - 74 y^2 + 1225 = 1792$

or $y^4-74 y^2 - 567=0$

or $(y^2-81)(y^2 + 7) = 0$ so $y^2=81$ as $y^2 \ge 0$

giving $y = \pm 9$ or $ x \in \{ 11,7\}$ the solution set  

2021/064) Solve in positive integer a,b $a^3-b^3 = ab + 11$

This type of problem is best solved by finding the bound and checking values in the bound

We have a=b is not a solution because LHS = 0 and RHS= 11

So $(a-b) >=1$ as $a > b$ because RHS is positive

Now $a^3-b^3 = (a-b)(a^2 + b^2 + ab)$

As and b are positive so AM, GM inequality gives   $a^2 + b^2 >= 2ab$

So we have $a^3 - b^3 >=  (a-b)(3ab)$

Or $a^3-b^3 >= 3ab$

So from the given relation $3ab < = ab +11 $ or $ 2ab <=11$ or $ab < 6 $ so $b<=2$ as   $a<=b$

b = 1 gives $a^3 = a + 12$ a =3 gives LHS = 8 RHS = 15

a = 4 gives LHS = 64 RHS = 16 so no integer solution LHS is greater than RHS 

b = 2 gives $a^3 = 2a + 11$ a =3 given LHS = 8, RHS = 17

a = 4 gives LHS = 64 RHS = 19 

So no solution

Hence there is no solution to this equation 

2021/063) Solve $x+\lfloor 2x\rfloor +\lfloor 3x\rfloor=7$

 Because RHS is integer so LHS is integer

As $\lfloor 2x \rfloor$ and $\lfloor 3x \rfloor$ are integers so x is integer so   $\lfloor 2x \rfloor = 2x $ and  $\lfloor 3x \rfloor = 3x$

so x + 2x + 3x = 6x = 7 so $x = \frac{7}{6}$ which is not integer so NO solution 

2021/062) Solve in integers x,y such that $2^x-3^y=7$

x and y have to be non -ve integers because otherwise we shall get a fraction

We have $3^y \equiv 0 \pmod 3$ for  y positive and and 1 for y zero

So let us take the 2 cases separately

$y = 0 => 2^x = 8$ or x = 3 so a solutions (x=3,y=0)

Now let us take y positive so we get

$2^x \equiv 1 \pmod 3$ so x has to be even as $2^x \equiv 1 \pmod 3$ for even x and -1 for odd x

so x = 2n.

Working in mod 4 we get $3^y \equiv 1 \mod 4$ so y has to be even as  $3^x \equiv -1 \pmod 4$ for even x and -1 for odd x

So we get y = 2m

Now $2^(2n) - 3^(3m) = 7$ or factoring we get

$(2^n+3^m)(2^n- 3^m) 7$ and as 7 is prime we have

$2^n+3^m = 7$ and $2^n-3^m = 1$

or solving we get $2^{nd}$ solution (n= 2, m = 1) or (x-4, y=2)

So 2 solutions are $(x,y) = \{(3,0), (4,2)\}$  

 

2021/061) Prove that the number 9999999+1999000 is composite.

Let x = 1000 to keep it simple

So we get

$9999999+1999000  = 10^7-1 + 1999 * 1000$

$= 10 * 10^ 6 - 1+ (2000-1)* 1000$

$= 10 x^2 - 1 + (2x-1) x$ putting 1000 = x

$=12x^2 - x - 1$

$= (4x+1)(3x-1)$

$= 4001 * 2999$ putting back x = 1000 

Hence it is composite 

2021/060) Find a polynomial of degree 3 with real coefficients such that each of its roots is equal to the square of one root of the polynomial $P(x)=x^3+9x^2+9x+9$.

Because the square of the roots shall be roots of the required polynomial so we must have $\sqrt x$ as roots of P(X)

So $(\sqrt x)^3 + 9 x + 9 (\sqrt x) + 9 = 0$

Or $(\sqrt x)(x + 9) = - 9(x+1)$

Or squaring $x(x+9)^2 = 81(x+1)^2$

Or $x(x^2 + 18 x + 81) = 81 x^2 + 162x + 81$

Or $x^3 - 63x^2 - 81 x - 81 = 0$

This is the required equation

2021/059) Let $p$ be a prime number. Find the fractional part of $\dfrac{(p+1)!}{p^2}$.

 Because p is prime by wilson theorem $p | (p-1)!+1$

Or $p^2| (p! + p)$

or $p^2| (p! + p) (p+1)$

Or  $p^2| (p+1)! + p(p+1)$

or $(p+1)!  \equiv  -p(p+1) \pmod p^2$

So fractional part of $\frac{(p+1)!}{p^2}$ is same as fractional part of $\frac{-p(p+1)}{p^2}$ or is $\frac{p-1}{p}$

2021/058) FInd integers a and b such that $5\frac{3}{a} * b\frac{1}{2} = 19$

it is easier to work for b rather than a  as the term $b\frac{1}{2}$ is between 2 integers

Now $ 5 < 5\frac{3}{a} < 6$ 

Now $ 5 * 4 = 20 > 19$ and $ 6 * 3= 18 < 19$ so $b\frac{1}{2}$ is between 3 and 4 and hence b = 3

$b\frac{1}{2} = \frac{7}{2}$

So  $5\frac{3}{a} = 19 * \frac{2}{7} = \frac{38}{7}= 5 \frac{3}{7}$

Or a = 3 

2021/057) Given $a^2+b^2=c^2+d^2=1$ and ac+bd=0.Compute the value of ab+cd

Because $a^2 + b^2 = 1$ we can take $a = \sin\alpha$ and $b = \cos \alpha $

Similarly $c=\sin \beta$ and $d = \cos\beta$

$ac + bd = \sin\alpha \sin  \beta + \cos \alpha \cos \beta = 0$

or $\cos (\alpha - \beta) = 0\cdots (1)$  (using formula of cos of difference)

$ab + cd = \sin \alpha \cos \alpha + \sin \beta \cos \beta = \frac{1}{2}(\sin 2 \alpha + \sin  2 \beta)$ (using formula for sin of twice angle)  

$= \frac{1}{2} \sin (\alpha + \beta) \cos (\alpha - \beta)$ using sum of sines

$= \frac{1}{2} \sin (\alpha + \beta) * 0$ using (1)

= 0