2026/016) Find all solutions of the linear congruence $3x−7y \equiv 11 \pmod {13}$

We have as 13 is a prime number we can choose any value of a x and the choose y in terms of x

We have $7y \equiv  3x -11  \pmod  {13}$

Now 7 needs to be multiplied by its inverse to get coefficient of y as 1 so multiplying by 2 (inverse of 7) we get

$y \equiv  6x - 22 \pmod  {13}$

Or   $y \equiv 6x-9  \pmod {13} $

For x = 0 to 12 mod 13 we get corresponding  value of y 

2026/015) Show that difference between squares of 2 conscutive triangular numbers is a perfect cube

We have $T_n = frac{n(n+1)}{2}$ 

To avoid fraction we have $2T_n = n(n+1)$

Now we have 

$ (2T_{n+1})^2 -   (2T_{n})^2 = ((n+1)(n+2))^2 -((n+1)(n+2))^2 $

Or $4(T_{n+1}^2-T_n^2) = (n+1)^2((n+2)^2 - n^2) = (n+1)^2 * 4 (n+1))= 4(n+1)^3$

Or   $T_{n+1}^2 -   T_{n}^2 = (n+1)^3$

 

2026/014) Find all integer N such that $\lfloor \sqrt{n} \rfloor$ divides n

 Any number from $k^2$ to $k(k +2)$ shall have square root between $k$ and $k+1$ as $k(k+2) +1$ shall have square root $k + 1$

So the floor function of any number $k^2$ to $k(k+2)$ shall be $k$. For $k$ to divide the number with in the range it has to be of the form $k^2$ or$ k(k+1)$ or $k(k+2)$

 So for any $k $ $k^2 , k(k+1),k(k+2)$ satisfy the criteria

2016/013) Show that the tens digit of $3^n$ is always even

We know working in modulo 20 as 3 is co-prime to 20 

$3^0  \equiv 1 \pmod {20}$

$3^1  \equiv 3 \pmod {20}$

$3^2  \equiv 9 \pmod {20}$

$3^3  \equiv 7 \pmod {20}$

$3^4  \equiv 1 \pmod {20}$

It repeats from this point,

As $3^n \pmod {20}$  is single digit so tens digit is even.

2026/012) The sum of two numbers is 667, the ratio of their LCM to GCF is 120. What are the two numbers?

 

Let $GCD(m,n) = k$ then we have

$m = ka$

$n = kb$

For some integers a and b where GCD(a,b) = 1

LCM = kab and GCD = k

So $ab = 120$

And $m +n = k(a+b)$

$m +n = 667 = 29 * 23$

So k can be $1,23,29, 667$

If $k =$1$  $a +b = 667$ and $ab = 120$ this is not possible

If $k = 667$ $a + b = 1$ so this is not possible

If $k = 23$ $a +b = 29, ab = 120$ so $a = 24,b =5$ or vice versa giving $(552,115)$  or $(115,552)$

if $k = 29$ $a +b = 23, ab = 120$ so $a = 15, b =8$ or vice versa giving$ (345,232)$ or $(232,345)$

So solution set is $\{(552,115), (115,552), (345,232), (232,325)\}$

2026/011) Solve in integer $x^2+4x +2 \equiv 0 \pmod 7$

 To complete square add 2 on both sides to get 

$x^2+4x +4 \equiv 2 \pmod 7$

or $(x+2)^2  \equiv 2 \pmod 7\cdots(1)$

now working mod 7

 $(0^2  \equiv 0 \pmod 7\cdots(2)$

 $(1^2  \equiv 1 \pmod 7\cdots(3)$

 $(2^2  \equiv 4 \pmod 7\cdots(4)$

 $(3^2  \equiv 2 \pmod 7\cdots(5)$ 

 from  (1) and (5) we have

 $(x+2)  \equiv 3 \pmod 7\cdots(6)$ or  $(x+2)  \equiv 3 \pmod 7\cdots(7)$ 

or $x\equiv 1 \pmod 7$ or    $x\equiv 2 \pmod 7$

 

 

 

2026/010) Is it true that $a^2+ab+b^2$ is divisible by 7 for infinitely many coprime pairs (a,b)

As a = 0 gives b= 0 so a = 7k and b= 7m satisfy the given equation but they ate not co-prime  

Because of symmetry let is assume $a=mb \pmod 7$ 

Putting in the given equation we get $m^2+m+1 \equiv 0\pmod 7$

As $m=1$  is not a solution to it multiplying by (m-1) on both sides   we get 

$m^3 \equiv 1 \pmod 7$ 

So $ m \equiv 1  \pmod 7$ or $ m \equiv 2  \pmod 7$ or $ m \equiv 4  \pmod 7$ but as $ m \equiv 1  \pmod 7$ is not a root of original eqution so solution is $ m \equiv 2  \pmod 7$ or $ m \equiv 4  \pmod 7$

So we can chose $(7p+m, 7q+2m)$ or $(7p+m, 7q+2m)$  p and q to be chosen such that the pairs form a co-prime

One set the infinite co-primes $(7p+1,7p+2)$ for any p