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 

 

 

 

 

2026/009) If $abc=1$ and $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ how can I show that a or b or c equals 1

We have $abc=1\cdots(1)$

 So $\frac{1}{a} = \frac{abc}{a} = bc$

Similarly 

$\frac{1}{b} = ca$

And 

$\frac{1}{c} =   ab$

Hence

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = bc+ca+ab\cdots(2)$

From above and

$a+b+c= \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

We have  

Hence $a+b+c = ab+bc+ca\cdots(3)$ 

a,b c are roots of equation

$P(x) = x^3-(a+b+c)x^2 +  (ab+bc+ca)x - abc=0\cdots(4)$

Let $a + b+c = m\cdots(5)$

So we get  from (4), (3), (5) 

$P(x) = x^3-mx^2+mx -1=0$ 

1 is a root of above equation as P(1) is zero

As a , b,c are roots one of $a,b,c$ is 1

Proved  

2025/008) Is it true that every odd number divides some number of the form $2^n−1$

 

Let the odd number be k. now let us take the number sequence $2^2–1, 2^3–1, 2^4–1$ so on k values that is upto $2^k-1$

Let us calculate all k remainder values when the above expressions are divided by k (as the remainder can be from zero to k-1) then we have on difference is zero and we are done.

If all k values are not different then we have say for 2 values of m and n the remainder must be same. without loss of generality let us assume $m > n$

So $(2^m-1) - (2^n-1) = 2^(m-n) -1 * 2^n$ is divisible by k. As k is odd $gcd(2^n,k) = 1$ so k must divide $2^(m-n) -1$

Note: The above is for the persons who are not familiar with number theory  or want to know from $1^{st}$ principle. 

n divides $2^{\phi(n)} -1$ as n is co-prime to 2 as per  Eulers theorem in number theory.

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