2026/004) If roots of the equation $a(b−c)x^2+b(c−a)x+c(a−b)=0$ are equal, then prove that 'a', 'b', 'c' are in harmonic proportion.

 Because roots are equal so the descrminamt is zero

or $b^2(c-a)^2-4ac(b-c)(a-b) = 0$

becuase we need to show that $a,b,c$ are in HP so $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP

now given relation is

$\frac{1}{q^2} (\frac{1}{r} - \frac{1}{p})^2- 4(\frac{1}{p})(\frac{1}{q})(\frac{1}{q}- \frac{1}{r}) (\frac{1}{p}- \frac{1}{q})=0 $ 

or $\frac{(p-r)^2}{p^2q^2r^2}$$ - 4(\frac{(r-q)(q-p)}{q^2r^2p^2})=0 $

or $(r-p)^2 -4 (r-q)(q-p) = 0$

Let $x= r-q$ and $y = q-p$

so we get $r - p = x + y$

or $(x+y)^2 - 4xy= 0$

or $(x-y)^2 = 0$

or $x = y$ or $q-p = r-q$ so   $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP and hence $a,b,c$ are in HP 

 

 

2026/003) Solve in positive integers $a^2=b^4 + 16b + 1$

 b has to even because if b is odd then we have RHS is of the form 4k +2 and it cannot be a perfects squares

now next square after $b^4$ is $(b^2+1)^2$ and the given expression can be a square(not necessary it will be) if greater than or equal to $(b^2+1)^2$

$b^4 + 16b + 1 >= (b^2+1)^2$

or $b^4 + 16b + 1 >= b^4 + 2b^2 +1 $

or $16b >= 2b^2$ or $ b <= 8$

So we need to try even values of b less or equal to 8 giving $b= 2 =>  a^2 = 49 =>a = 7$ 

$b =4 => a^2 = 256 + 64 +1 = 321$ no solution

$b = 6=> a^2 = 1395$ no solution

$b = 8 => a = b^2+1 = 65$

Solution $(7,2)$ and $(65,8)$

2026/002) Let a,b,c be integers. If $4a+5b−3c$ is divisible by $19$, prove that $ 6a−2b+5c$ is also divisible by $19$

 As 4 and 6 have LCM 12 so we should multiply by 3 and proceed

$4a + 5b -3c$ is divisible by 19 so multiply by 3 to get $12a + 15b -9c$ is divisible by 19. adding $19c-19b$ we get

$12a - 4b +10c$ is divisible by 19

or $2*(6a-2b + 5c)$ is divisible by 19 and as 2 is co-prime to 19 hence $6a-2b+5c$  is divisible by 19

2026/001) How do I find the smallest positive integer n, such that each digit of n is either 0 or 7, and n is divisible by 15?

It is divisible by 15. So it is divisible by both 3 and 5

Because it is  divisible by 5 so unit digit is 0 or 5, As we have to chose from 0 and 7 so it is zero. Now we cannot have a zero in other position as it will increase the value of the number as it is additional digit and does not serve any other purpose.Sso other digits shall be 7 and we require 3 7s to be divisible by 3 or the number as 7770( 15 * 518) .

 

2025/034) If p can be expressed as sum of 2 squares say $p = x^2+y^2$ then express 2p ,5p, 19p as sum of 2 squares

If p can be expressed as sum of 2 squares as below

$p =  x^2+y^2$

We know $2= 1^2 + 1^2$

Now using the identity 

$(a^2+b^2)(x^2 + y^2) = (ax+by)^2 + (bx-ay)^2$ (we can expand and see the result

We get $2p = (x+y)^2+ (x-y)^2$

Further $5 = 2^2 + 1^2$

So we get $5p = (2x+y)^2 + (x-2y)^2$

But 19 is of the form 4n + 3 and it cannot be expressed as sum of 2 squares so 19p cannot be expressed as sum of 2 squares  

 

 

2025/033) Let a,b,c be the consecutive sandwiched integers between any cousin primes. Is it true that either $5∣(a+b+c)$ or $5| (a^2+b^2+c^2)$ ?

 Let x,y be cousin primes. We have x,a,b,c,y are 5 numbers.

now a = b-1 and c = b +1

so We have $a+b+c = 3b$

$a^2+b^2+c^2 = (b-1)^2 + b^2 + (b+1)^2 = 3b^2 + 2$

x is a prime so x cannot be multiple of 5 or b cannot be of the form 5k+2

y is a prime so y cannot be multiple of 5 or b cannot be of the form 5k +3

If b is of the form 5k $a+b+c$ is divisible by 5

If b is of the form 5k+ 1  $a^2+b^2+c^2 = 3(5k+1)^2 + 2 = 3(25k^2+10k+1) + 2 = 75k^2+ 30k + 5 = 5(15k^2+ 6k+1)$ multiple of 5

 If b is of the form 5k+ 4  $a^2+b^2+c^2 = 3(5k+4)^2 + 2 = 3(25k^2+40k+16) + 2$

$= 75k^2+ 120k + 50 = 5(15k^2+24k+10)$ multiple of 5

 So either of them is multiple of 5 

 

  

2025/032) How do I find the roots of the equation $x^3-14x^2+56x–64=0$, the roots are in geometric progression?

 roots are in GP so sw have root are $r, \frac{r}{a},ra$ so we get product of roots 

=$r^3= 64$ or $r=4$ So one root is 4 and hence (x-4) is a factor.  

we have by synthetic division.

$x^3-14x^2+56x–64=0= (x-4)(x^2-10x + 16)$

one factor is (x-4) and other factor is quadratic  and can be factored as $(^2-10x+16=(x-2)(x-8)$

giving roots 2,4,8 and they are in GP(just for cross checking)