2026/60) Show that $f(n)=n^5+n^4+1$ is not prime for $n>1$

We shall prove a stronger result. we shall show that

$g(n) = n^{3k+2} + n^{3m+1} + 1\cdots(1)$ is not a prime for $n>1$ and $k+m > 0$

To prove it we shall snow that is is divisible by $n^2+n+1$

We have 

$n^2+n+1 = (n-\omega)(n-\omega^2)..\cdots$  where $\omega$ is cube root of 1

And $w^3 = 1\cdots(3)$

And $w^2+w+1=0\cdots(4)$ 

We shall snow that g(n) is divisible by  $ (n-\omega)$ and $ (n-\omega^2)$

 We get putting $\omega$ in (1) for n 

$g(\omega) =   \omega^{3k+2} + \omega^{3m+1} +1$

$= (\omega^3)^k \omega^2 +   (\omega^3)^m \omega + 1$

$=   \omega^2 +   \omega + 1 = 0$ using (3) and (4)

So $g(n)$ is divisible by  $(n-\omega)$

 Similarly $g(n)$ is divisible by $(n-\omega^2)$

So $g(n)$ is divisible by  $n^2+n+1$

Because  $k+m$ is greater than 0 so at least one of them is greater than zero

So $g(n) > n^2+n+1$ and as $n^2+n+1 > 0$ g(n) is product of 2 numbers neither is 1 

Hence g(n) is composite

 Putting k = m = 1 we get $f(n)$

So f(n) is not prime 

 Proved

 

2026/059) Show that the product of two positive integers of the form $a^2+ab+b^2$ has the same form.

Let two integers be $a^2+ab+b^2$ and $c^2+cd+d^2$

We have

$a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)\cdots(1)$ where $\omega$ is complex cube root of 1

And

$c^2+cd+b^2 = (c-d\omega)(c-d\omega^2)\cdots(2)$ 

 Multiplying (1) by (2) we get

 $(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$

Now as $\omega$ is cube root of 1 so we have

$\omega^2+\omega+1 = 0$

Or 

$\omega^2 = - (1+\omega)\cdot(3)$

Also

 $\omega = - (1+\omega^2)\cdot(4)$ 

Now get us calculate $(a-b\omega)(c-d\omega)$

$(a-b\omega)(c-d\omega) = ac -\omega(ad + bc) + bd\omega^2$

$= ac -\omega(ad+bc) -bd(1+\omega)$ from (3)

$= (ac-bd) -\omega(ad+bc+bd)\cdots(5)$

 

Now get us calculate $(a-b\omega^2)(c-d\omega^2)$

$(a-b\omega^2)(c-d\omega^2) = ac -\omega^2(ad + bc) + bd\omega^4$

$= ac -\omega^2(ad+bc) +bd(\omega)$ as $\omega^3=1$ 

$= (ac -\omega(ad+bc) +bd(1+\omega^2)$ from (4)

 $= (ac-bd) -\omega^2(ad+bc+bd)\cdots(6)$

Now   $(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$

$=(a-b\omega)(c-d\omega)(a-b\omega^2)(c-d\omega^2)$

$=((ac-bd) -\omega(ad+bc+bd))((ac-bd) -\omega^2(ad+bc+bd))$

$=((ac-bd)^3 +(ac-bd)(ad+bc+db)+(ad+bc+db)^3$                               using $x^3+y^3=(x-\omega y)(x-\omega^2 y)$

= $m^2+mn + n^2$ where  $m= ac-bd$ and $n = ad+bc+bd$

Hence proved  

  

 

2026/058) Find the number of ordered triples of positive integers $(a,b,c)$ such that $6a+10b+15c=3000$

We are given

$6a+10b+15c=3000\cdots(1)$ 

We have rearranging the terms let us get a in terms of others 

$6a = 3000 - 10b-15c = 5(600-2b-2c)$

As the RHS is divisible by 5 and GCD(5,6) = 1 so a is divisible by 5 and so

$a=5x\cdots(2)$ for some x

Again to get b in terms of others

 $10b = 3000 - 6a-15c = 3(1000-2a-5c)$

As the RHS is divisible by 3 and GCD(3,10) = 1 so b is divisible by 3 and so

$b=3y\cdots(3)$ for some y

 Again to get c

 $15c = 3000 - 6a-10b = 2(1500-3a-5b)$

As the RHS is divisible by 2 and GCD(2,15) = 1 so c is divisible by 2 and so

$c=3z\cdots(4)$ for some z

Putting the values a,b,c from (2),(3),(4) respectively in (1)
 

$30x + 30y + 30z = 3000$

Or $x + y + z = 100$

We need to find x,y,z all 3 positive integers sum is 100

Let us assume that there are 100 stones in a line . There are 99 gaps. we can make them into 3 parts of each part containing above one by putting 2 sticks in 99 gaps. this can be done in $99 \choose 2$ ways

  

 

 

2026/057) Find the smallest positive integer n such that $3^5 | 10^n-1$

Before finding the n let us make a few obsevations

We have for any n $9 | 10^n-1$

That is   $3^2 | 10^n-1$

And as $ 3| 111$ or $3 | \frac{10^3-1}{9}$

Hence $3^3 | 10^3-1$

Now we need to move from $3^3$ to $3^5$

999 is divisible by 27

Appending another 999 to the left shall give 999999 and it is divisible by 27 appending 9 or 99 does not make if divisible by 27

But appending 999 to the left is adding $999 * 10^3$

As $243 = 27 *9$ we need to keep a-dding until the  

Let us keep appending say n terms

We get $999(1+ 10^3 + 10^6 + \cdots 10^{3n})$

For this to be divisible by 243 as 999 is divisible by 27  $1+ 10^3 + 10^6 + \cdots 10^{3n})$ must be divisible by 9.

Each term of   $1+ 10^3 + 10^6 + \cdots 10^{3n})$ divided by 9 leaves a remainder 1 so there should by 8 terms and so n= 24

Then we get the number = $999(1+10^3 + \cdots 10^24)$

$= (10^3-1)(1+10^3 + \cdots 10^24)= 10^27-1$

so n= 27 

  

 

 

 

 

 

2026/056) Prove that 1280000401 is composite.

 We have 

$1280000401 = 1280000000 + 400 + 1= 20^7 + 20^2+ 1$

This is $x^7 + x^2 + 1$ where x = 20

All the terms have coefficient 1

There is an exponent 7 and and an exponent 2  one if of the form 3n + 2 and another of the form 3n+1

This is divisible by $x^2+x+1$ hence composite

Let us prove the same.

$x^7 + x^2 + 1$ 

$= x^7 -x + x^2 + x + 1$

$=x(x^6-1) + x^2 + x + 1$

$=x(x^3+1)(x^3-1) + (x^2 + x + 1)$

$=x (x^3+1)(x-1)(x^2+x+1)+ x^2 + x + 1$

$=(x^2+x+1) (x(x^3+1)(x-1) + 1$

We have factored the same and 

so the given  number is composite

 

 

 

 

2026/055) show that 1994 devides $10^{900} - 2^{1000}$

Now Let us factor 1994

$1994 = 2 * 997$

We need to show that it is divisible by 2 and 997 

RHS is even so is is divisible by 2

We need to show that the RHS is divisible by 997 

We have as 997 is close to 1000 so let is brrin power of 10 and 2 to be as close as 997

$10^3 \equiv 3 \pmod {997}\cdots(1)$

and  $2^{10} \equiv 27 \pmod {997}\cdots(2)$

so $10^{900} - 2^{1000}$

 $=({10^9})^{100} - (2^{10})^{100}$ )

This is divisible by $10^9 - 2^{10}$

Now $10^9 - 2^{10}\pmod {997}$

$= (10^3)^3 - 2^{10}\pmod {997}$

 $= 3^3 - 27 \pmod {997}$

$= 27-27 \pmod {997}$

= 0

So the number is divisible by 997

As it is divisible by 997 and 2 so it is divisible by 1994

 

 

 

 

2026/054) What is the next number of the sequence 256, 1156, 4356..

The sequence is the squares of numbers which leaves a remainder 56 when divided by 100 that is

$n \equiv 56 \pmod {100}$ 

We can look at the numbers mod 100 and then find the numbers from table lookup. However this is not interesting

To keep the things simple let us find the numbers mod 10 and then make it to numbers mod 100.

Let us look at the number mod 10 and find the square mod 10 we get (1st is number mod 10, 2nd is number square mod 10 $(0,0),(1,1),(2,4),(3,9),(4,6),(5,5),(6,6),(7,9),(8, 4),(9,1)$

So then umber is 4 mod 10 or 6 mod 10

That is number is of the form 10n+ 4 or 10n+ 6

Now we need to find n in both the forms.  

We can take the x as 2 digit number . adding 100 to x shall not change the 2 digits mod 100

This is because

 $(100n + k)^2 = 10000n^2 + 200nk + k^2$

so  $(100n + k)^2 = k^2\pmod {100}$

Consider first form $x = 10n + 4$

Now $(10n+4)^2  = 100n^2 + 80n + 16$

so $80n + 16 = 56\pmod {100}$

or $80n = 40 \mod 100$

dividing by 20 on both sides

$2n = 1\pmod 5$

or $n = 3 \mod 5$ or $n \in \{3,8\}$ as we are looking for single digit number

So the number is $34 \pmod {100}$ or $84 \pmod {100}$

we can combine both to get $x \equiv 34\pmod {50}$  

Considerfnext form $x = 10n + 6$

Now $(10n+6)^2  = 100n^2 + 120n + 36$

As remainder is 56 

so $20n + 16 = 56\pmod {100}$

or $20n = 20 \mod 100$

dividing by 20 on moth sided

$n = 1\pmod 5$

or $n = 1 \mod 5$ or $n \in \{1,6\}$ as we are looking for single digit number

So the number is $16 \pmod {100}$ or $66 \pmod {100}$

So we have  $x \equiv 16 \pmod {50}$ or $x \equiv 34 \pmod {50}$