2022/026) Let P(x) be any polynomial with integer coefficients such that P(21)=17,P(32)=−247and P(37)=33. Prove that if P(N)=N+51 for some integer N, then N=26.

Using the property of polynomial that $a-b | p(a)- p(b)$ we have between N and 21

$N-21 | P(N) - p(21)$

As $P(N) = N+ 51$

We get $N-21 | N+ 51 -  17$

or $N-21 | N + 34| or | N - 21 |(N+ 34) - (N-21) $

or $N-21 | 55$

This gives N-21 = 1 or 5 or 11 or 55 or -1 or -5 or - 11 or -55

So N = 26 or 32 or 36 or 76 or 20 or 16 or 10 or - 34 (1)

Now with 32 we get $N- 32 | N + 51 - (-247)$ or $N-32 | 330$ (2)

Finally with 37 we get $N -37 | N + 51 - 33$ | or $N- 37 | 55$

from  set of (1) we get 26 or 32 or 36 

from above (3) values we get   N = 26

so N  = 26 is the answer

 

2022/025) if $a-\frac{1}{a} = b$, $b-\frac{1}{b} = c$, $c-\frac{1}{c} = a$ evaluate $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$

 We are given $a-\frac{1}{a} = b\cdots(1)$

$b-\frac{1}{b} = c \cdots(2)$

$c-\frac{1}{c} = a\cdots(3)$

From (1)$\frac{1}{a} = a-b $

dividing both sides by b

$\frac{1}{ab}  =  \frac{a}{b} - 1  = a(b-c)  -1 $ using |(2)

similarly

$\frac{1}{bc}  =  \frac{b}{c} - 1  = b(c-a)  -1 $

and $\frac{1}{ca} = \frac{c}{a} - 1 = -c(a-b) = 1 $

adding above 3 we have $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}= - 3 $

2022/024) Given $a_1^2+5a_2^2=10,\,a_2b_1-a_1b_2=5$ and $a_1b_1+5a_2b_2=\sqrt{105}$ for $a_1,\,a_2,\,b_1,\,b_2\in R$, evaluate $b_1^2+5b_2^2$.

we have

$(p^2+q^2)(l^2 + m^2) = (pl-qm)^2+ (pm + ql)^2$




putting $p= a_1$, $q = \sqrt{5} a_2$ ,$l= b_1$, $m = \sqrt{5} b_2$



we get $(a_1^2 + 5a_2^2)(b_1^2+5b_2)^2 = (a_1b_1 + 5a_2b_2)^2+5(a_1b_2 - a_2b_1)^2$



putting values from given conditions we get



$10(b_1^2+5b_2^2) = 105 + 5 * 5^2$

or $10(b_1^2 + 5b_2^2) = 230$

or $b_1^2 + 5b_2^2 = 23$

2022/023) Solve for x $\sqrt{3-x} - \sqrt{1+x} >= \frac{1}{2}$

We first note that $3-x \ge 0$ and $1+x <=0$ as both are under square root and so we have $-1 \le x \le 3$

 Secondly we have the LHS is monotonically decreasing as x increases

so let uf have $f(x) = \sqrt{3-x} - \sqrt{1+x}$

We have the LHS is monotonically decreasing as x increases and if is defined in $[-1,3]$

now $f(-1) = \sqrt{3-(-1)} - \sqrt{1+{-1}} = 2$ and $f(3) = \sqrt{3-3} - \sqrt{1+3} = -2$

so we need to solve for  $\sqrt{3-y} - \sqrt{1+y} = \frac{1}{2}$

and then $y \le x \le -3$ 

we need to solve 

$\sqrt{3-y} - \sqrt{1+y} = \frac{1}{2}$

square both sides to get $3 - y + 1 + y - 2\sqrt{(3-y)(1+y)} = \frac{1}{4}$

or $4 - \frac{1}{4} = 2\sqrt{(3-y)(1+y)}$

square both sides to get $\frac{225}{64} = (3-y)(1+y)$

or $\frac{225}{64} = (3-y)(1+y)$

putting y = z + 1 we get $\frac{225}{64} = (2-z)(2+z) = 4 - z^2$

or $z^2 = \frac{31}{64}$

so $z = \pm \frac{\sqrt{31}}{8}$

so $y = 1 \pm \frac{\sqrt{31}}{8}$

now whether we show take the plus or minus we check  that $f(1) = \sqrt{2} - \sqrt{2} = 0$ wich does not sattisfy the condition so we must have -ve numebr and hence

$y =    1  - \frac{\sqrt{31}}{8}$

or $-1 \le x \le  1  - \frac{\sqrt{31}}{8}$

or 

2022/022) Find all solutions in positive integers x, y, z of the equation $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 3$

We have by AM GM inequaity among $\frac{x}{y}$ ,$\frac{y}{z}$, $\frac{z}{x}$

we have 

$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} >= \sqrt[2]{\frac{x}{y} * \frac{y}{z} * \frac{z}{x}}$

 or  $\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} >= \sqrt[2]{1}$

or  $\frac{x}{y} + \frac{y}{z} + \frac{z}{x}>=  3$

this is greater than 3 if all are not equal and is 3 if all are equal

or  $\frac{x}{y} = \frac{y}{z} = \frac{z}{x}$

or x = y = z (any integer)  

2022/021) find rational x,y z, such that $x^2 + y^2 + y^2 + x + y + z =1

 we have 

$x^2 + y^2 + z^2 + x + y + z = 1$

or $x^2 + x + y^2 + y + z^2 + z = 1$

we shall complete the square so let us multiply 4 to avoid fraction

$(4x^2 + 4x ) + (4y^2 + 4y) + (4z^2 + 4z) = 4$

adding 1 to each term on the left

 $(4x^2 + 4x + 1) + (4y^2 + 4y + 1) + (4z^2 + 4z+ 1) = 7$

or $(2x+1)^2 + (2y +1)^2 + (2z+1)^2 = 7$

multipying by LCM of denomiantor of 2x+1,2y+1,2z+1 we get

$p^2 + q^2 + r^2 = 7m^2$ for integers p,q,r,m.

this has a least m which has a solution

let m be even say 2n

then  all a,b,c have to be even because RHS is divisible by 4 and if any one to 3 numbers is odd then working in mod 4 and summing we get a remainder of 1to 3 when divided by 4.

so all are even

so m cannot be even

so m is odd.

because m is say m = 2k+1

so $7m^2 = 7(2k+1)^2 = 7 * 4k^2 + 7 * 4k + 1$

ot $7m^2 = 7 * 4k(k+1) + 7$

which leaves a remainder 7 when divided by 8

a number if odd on squaring leaves a remainder 1 when divided by 8 and if even then remainder is 0

sum of 3 such numbers (0 or 1) cannot be 7.

so the equation has has no solution and hence original equation has no solution.

   

2022/020) Given $m^2 = n +2 $, $m^2= n + 2$ and m and n are unequal Compute $4mn - m^3 - n^3$

we are given 

$m^2 = n +2 \cdots(1)$

$n^2 = m + 2\cdots(2)$

$4 mn - m^3 - n^3$

$= 4mn - m(m^2)  - n(n^2)$

$= 4mn - m(n+2)  - m(n+2)$ using (1) and (2)

$= 2mn - 2(m+n)$

or $4 mn - m^3 - n^3=  2mn - 2(m+n)\cdots(3)$ 

we need to commte mn and m + n

sutarcting (2) from (1)

$m^2 - n^2 = n - m$

as n-m is not zero divding by n-m we have

$m+n = -1\cdots(4)$

adding (1) and (2)

$m^2 + n^2 = (m+n) + 4 = -1 + 4 = 3\cdots(5)$ putting value of m + n from (4)

or

Hence $2mn = (m+n)^2 - (m^2+n^2) =  1- 3 = -2$

putting the values from (4) and (6) in (3) we get  

 $4 mn - m^3 - n^3=  2mn - 2(m+n) =  -2 -2 (-1) = 0$ 

 

2022/019) Find real x and y such that $16^{x^2+y} + 16^{y^2 + x} =1$

 We have both $16^{x^2+ y}$ and $16^{y^2+x}$ positive.

So we can apply AM GM inquality gettting

$\frac{16^{x^2+y} + 16^{y^2+x}}{2}\ge (16^{x^2+y} 16^{y^2+x})^{\frac{1}{2}}$

or $16^{x^2+y} + 16^{y^2+x}\ge  2 (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$

we are given LHS = 1 so 

$ 1 \ge 2 (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$

or $ 1 \ge 16^{\frac{1}{4}}  (16^{x^2+y+ y^2+x})^{\frac{1}{2}}$

or $ 1 \ge  (16^{x^2+y+ y^2+x+ \frac{1}{2}})^{\frac{1}{2}}$

or $x^2 + y + y^2 + x + \frac{1}{2} <=0$ 

Now we combine line terms and complete square to get 

$(x^2 + x + \frac{1}{4}) + (y ^2 + y + \frac{1}{4}) <= 0$

or $(x+\frac{1}{2})^2 + (y + \frac{1}{2})^2 <=0$

it is sum of 2 squares so cannot be be -ve so

both terms and sum has to be zero givng $x = y = \frac{-1}{2}$

2022/018) if a,b,c are roots of equation $x^3-24x^2+183x -440=0$ find the area of the triangle whose sides are a,b,c.

 Note I picked the question at https://www.youtube.com/watch?v=iyX-q3D91Ck&t=474s and here provide a diiferent solution

Let $P(x)= x^3-24x^2+183x -440\cdots(1)$

as a,b,c are roots of the equation  P(x) = 0 so we have following 2 observations

$P(x) = (x-a)(x-b)(x-c) \cdots(2)$

and comparing oefficient of $x^2$ we have

$a+b+c = 24\cdots(3)$

Now area of the triangle say A = $\sqrt{s(s-a)(s-c)(s-c}$

where s is semiperimeter  of triangle 

so $s=\frac{a+b+c}{2}= \frac{24}{2}$ putting value of a+b+c from (3)

or $s= 12\cdots(4)$ 

from (1) putting s for x we get $P(s) = (s-a)(s- b)(s-c)$

So A = $\sqrt{s(s-a)(s-c)(s-c)} = \sqrt{sP(s)}$

so using (1) and above

$A^2 = sP(s) = s(s^3 - 24s^2 + 183 s - 440) $

putting the value of s = 12 from (4) we get

$A^2= 12 *(12^3- 24 * 12^2 + 183 * 12  - 440) = 12(12^3 - 2 * 12^3 + 183 * 12 - 440)$

Or $A^2 = 12(- 12^3 + 183 * 12 - 440) = 12(- 1728 + 2196 - 440) = 12 * 28 = 336$

or $A = \sqrt{336} = \sqrt{4^2 * 21} = 4 \sqrt{21}$  

2022/017) find positive integer n such that $n^2-10n -22$ is same as product of digits of n

Let us fix the lower bound of n

now $n^2 -10n - 22 = (n^2 - 10n + 25) - (22+25)$ adding 25 for completing the square

$= (n-5)^2 - 47 > = 0$ as it has to be positive

so $(n-5)^2 >= 57$

or $(n-5) >= \sqrt{47}$ 

or $ n > 5 + \sqrt{47}$

as n is integer so n > 5 + 6$ as $6 < \sqrt{47} < 7$ so 5 + 6 is the largest value which does nt satisfy the condition

so $n >11$

now  n = 12 this gives product of digits = 2 and $n^2 -10n -22 = 12^2 - 10 * 12 - 22 = 2$ so they are same

so 12 is a sloution

for $n  > 12$ we have $22 < 2n$ and hence $n^2 - 10n - 22 > n^2 - 10n - 2n$

or $n^2 - 10n - 22 > n(n-12)$ and as $n-12$ is positive we have

$n^2 - 10n -22 >n$ for $ n > 13\cdots((1)$

We shall prove a little lemma:

For for any n the product of the digits of  n is less then n

proof:

let n be a k digit number starting with a.

the value of $n >= a* 10^{k-1}\cdots(2)$

the number that starts with a shall give the largest product when all rest (k-1) digits will be ragest that is 9

so product of digits  $ < a * 9^{k-1}\cdots(3)$

using (2) and (3) we get product of digts < the value n < value of exptession.

end of lemma

using the lemma and (1) we have for $n >12$ it is not possible for the expression . same as product of digits.

 

   

2022/016) find integers m and n such that $2^m - 2^n = 2016$

 we have clearly $m > n$

taking $2^n$ common we have $2^n(2^{m-n} - 1) = 2016$

clearly $2^{n}$ is even and $2^{m-n}-1$ is odd 

so let us factor 2016 to be a power of 2 multiplied by odd

$2016 = 32 * 63$

so n= 5 and $63 + 1 = 64= 2^6$ which is a power of 2 

so m -n = 6 or m = 11

so solution m= 11, n = 6 

2022/015) What are the prime numbers P that make 37P+4 the square of a number?

if is a square let it be $x^2$ 

$37p + 4= x^2$

or $37p = (x^2-4)  = (x-2)(x+2)$

as p is prime so one number is 37 and another is prime or one number is 37p and another number is 1

if x -2 = 1 then x +2 is 5 so it is not 37p

difference between x -2 and x+ 2 is 4 so numbers are 37 and 41 or 37 and 33.

but 33 is not prime so p = 41