2025/023) Solve $\sqrt{(x+2)} > x $

It is risky to square both sides as it shall give erroneous root.

Let us first find the range of x

as we can take the square root of a non -ve number only so we 

$x \ge -2$ 

for $x \in [-2,0] LHS is positve and RHS is -ve or zero. so  this is range in  -ve

To find the range is positive we square both sides to get 

$(x+2) > x^2$

or   $x^2-x-2 <0$

or $(x-2)(x+1) <0$ or   $x \in [-1,2)$

as x is positive so   $x \in [0,2)$

so combining we get  $x \in [-2,2)$ 

 

2025/022) How do you prove that $a^5−a$ is divisible by 10 (a is a natural number)

 Both a and $a^5$ are even or odd so $a^5-a$

Now if a is multiple of 5 $a^5$ is multiple of 5. so $a^5-a$ is divisible by 5

If a is not multiple of 5 then as per FLT $a^5 \equiv a \pmod 5$ so  $a^5-a$ is divisible by 5

In both cases  $a^5−a$ is divisible by 10

 

 

 

2025/021) Find integer solutions of the equation $2x - 3y = 8$

we can solve by fining GCD of 2 and 3 but here I shall solve it by a different approach 

We can rewrite this as

$3y = 2x -8$

RHS is even so LHS should be even or y should be even

So let $y =2k$

We get $6k = 2x -8$

Or x = $3k + 4$

So $x = 3k + 4$ and $y = 2k$ are parametric form of solution

by choosing any integer for k we can get particular solution

k = 1 gives (7,2)

k =2 gives (10,4)

2025/020) Given $ab = 10$ and $a- b= 3$ find $a^3-b^3$

We can solve the 2 equations and get $a=5$ and $b=2$ and hence $a^3-b^3=5^3-2^3=117$

but the above approach is longer as compared to the approach below

We have

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

or $a^3-b^3= (a-b)^3 + 3ab(a-b)$

Putting the values we get $a^3-b^3=3^3 + 3 * 10 * 3 = 117$

In this process we have avoided solving of equations  

    

2025/019)Find prime p such that 16p+1 is a perfect cube

$16p+1$ is perfect cube so let is be $x^3$

So $x^3-1 = 16p$

Or $(x-1)(x^2+x+1) = 16p$

Now $x^2+x+1$ is is odd so $16| x-1$

So $(x-1) = 16m$ and $x^2+x+1 = n$ where m and n are odd

So p = mn

As p is odd $m = 1, n = p$  or $m = p, n= 1$

n= 1 gives $x^2+x+1 = 1 $ or x = 0 which is not possible

so m= 1 and x= 17 giving n = 307 wihich is a prime

so $p = 307$    

 

 

2025/018) If a , b , c are any three integers then show that $7 | abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$

Le $P(a,b,c) =  abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$

 

Let us first find out  $a^3 \mod 7$

We have $a^3 \mod 7 \in \{1,-1\}$ when $a \mod 7 \ne 0$

if  $a \mod 7 = 0$ or $b \mod 7 =  0$ or $c \mod 7 \ne 0$ we have $7 |  P(a,b,c)$

$a^3 \mod 7 \in \{1,-1\}$

$b^3 \mod 7 \in \{1,-1\}$

$c^3 \mod 7 \in \{1,-1\}$

each can take one of two values and there are 3 differences so one of them has has to be zero

hence  $7 |  P(a,b,c)$

Proved  

    

2025/017) Solve in positive integers $x^2-xy+y^2 = 13$

we have 

$x^2-xy+y^2 = 13$

we can complete square by addition of some thing from $y^2$ to $x^2-xy$

to get $(x^2-xy + \frac{y^2}{4}) +  \frac{3y^2}{4} = 13$

or $(x-\frac{y}{2})^2 +  \frac{3y^2}{4} = 13$

or multiplying by 4 we get $(2x-y)^2 + 3 y^2 = 52$

as it is sum of positive numbers we get $3y^2 <=52$ or $ y <=4$'

$y =1$ gives $(2x-y)^2 = 49$ or $2x-y=7$ as -7 shall give -ve x so x = 4, y= 1

$y=2$ gives $(2x-y)^2= 40$ not a square

$y=3$ gives $(2x-y)^2 = 25$ or $2x-y =  5$ giving x = 4 , y = 3 and $2x-y = -5$ gives -ve x

$y=4$ gives (2x-y)^2 = 4 $, $2x-y=2$ gives x = 3, y = 4

 $2x-y=-2$ gives x = 1, y = 4

So we have solutions (4,1),(4,3),(1,4),(1,3)