2025/007) Can two different primes $p\ne q$ satisfy $p^q−q^p=1$

Because difference is one one of the term is even and other is odd . So either p or q (being prime) is 2.

Let p = 2

so $2^q-q^2$, $q =3$ gives $-1$,$ q = 5$ gives $7$ and this keeps going increasing so no solution

Now let us check with $q = 2$ 

So $ p^2 -2^p$  $p =3$ satisfies$ p = 5$ gives $-7$ and larger p gives smaller value decreasing so no more solution

So only solution $p =3, q = 2$

2026/006) Prove that for real $x_1,x_2,x_3$ $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$

We have

 $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)$

$=(x_1^2- \frac{4}{3}x_1x_2) + ( x_3^2-  \frac{4}{3} x_2x_3)+x_2^2$  Combining $x_1x_2$ with $x_1^2$ term and   $x_3x_2$ with $x_2^2$ term 

$=(x_1^2- 2 *\frac{2}{3}x_1x_2 + (\frac{2}{3}x_2)^2) + ( x_3^2-  2* ( \frac{2}{3} x_2x_3+  (\frac{2}{3}x_2)^2)+(x_2^2-  (\frac{2}{3}x_2)^2) - (\frac{2}{3}x_2)^2)$  completing the square and subtracting the same 

$=(x_1 - \frac{2}{3}x_2)^2 + ( x_3- \frac{2}{3}x_2)^2+\frac{1}{9} x_2^2$

This is positive or zero

So 

  $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)>=  0$

Hence  $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$ 

2026/005) How do I the smallest four-digit integer N, such that $\sqrt{3\sqrt{N}}$ is an integer?

We have one 3 under the square root so we should have another 3 that should be part of $\sqrt{N}$ So it should be $N= (3x)^2$ and x has to be a square so smallest 4 digit number giving $N= 9x^4$ satisfying that it is smallest 4 digit number    when $x=4$ giving  $N= 9 * 256$ or 2304 

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) .