2022/047) Solve in integers $y^2 - 2001 = n!$

 We have 2001 is divisible by 3 and not 9. And if n Is above 5 then n! is divisible by 9. So $2001 + n!$ is divisible by 3 but not 9. so it cannot be a perfect square. so checking from 0 to 5 we get y = 45 and n =4

2022/046) Find all n such that $7 | 2^n-1 $ and show that there is no positive n such that $7 | 2^n + 1$ (IMO 1964 problem 1)

because 7 is a prime so as per Fermats Little theoren 

$ 7 | 2^6-1$

now as $2^6$ leaves remainder 1 after dividing by 7 so it may be taht for some factor a of 6 $7 | 2^a-1$

we need to check for 1,2,3

so see $2^1-1 = 1$ $2^2-1 = 3$ and $2^3 - 1 = 7$ out f these 3 3 satisfies

as so k = 3m for all m satisfies.

Further $2^1+1 = 3$ $2^2+1 = 5$ and $2^3 +1 = 9$ out f these 3 none is divsible by 7 so there is no n such that $ 7 | 2^n+1$ 

proved 

2022/045) Find x given $(2+\sqrt{3})^x + (2-\sqrt{3})^x = 4$

 we see that $(2+\sqrt{3})(2-\sqrt{3}) = 4 - 1 = 1$

so $\frac{1}{(2+\sqrt{3}} = 2 - \sqrt{3}$

Let $(2+\sqrt{3})^x = y$ so $(2-\sqrt{3})^x = \frac{1}{y}$

so $y +\frac{1}{y} = 4$

or $y^2 - 4y + 1 = 9$

ot $y = \frac{4 \pm \sqrt{16-4}}{2}$

or $y= 2 \pm \sqrt{3}$

If we take $y = 2 + \sqrt{3}$ then $(2+\sqrt{3})^x = 2 + \sqrt{3}$ or $x= 1$

 Taking $y = 2 - \sqrt{3}$ then $(2+\sqrt{3})^x = 2 - \sqrt{3}$ or $x= -1$

so $x \in \{ 1, -1\}$

2022/044) If p, q, r, s are consecutive terms of an arithmetic sequence with common difference n how do you prove that $pqrs + n^4$ is a perfect square?

 Without loss of generality set n = 2x ( x need noy be integer so we can choose  

p = m - 3x, q = m - x, r = m + x, s = m + 3x

now $pqrs+n^2 = (m-3x)(m-x)(m+x)(m+3x)  + 16x^4$

$= (m- 3x)(m+ 3x)(m-x)(m+x) + 16x^4$ 

$= (m^2-9x^2)(m^2- x^2) + 16x^4$

$= m^4 - 10x^2m^2  + 9x^4 + 16x^4$

$=m^2 - 10x^2m + 25x^4 = (m^2 -5x^2)^2$ so it is a perfect square 

as it is integer square root is also integer 

2022/043) Sum of 13 numbers is 1000 what is the minimum LCM

The sum of 23 numbers is 1000 so the numbers can be 12 numbers 77 and 76, so the LCM>=77.

 the LCM is lower if 13th number is a factor of 12 numbers (which are same)

let us look at factors of 1000. and one factor is minumum 13.

we have 50 * 20. 7 numbers  can be 100 and 6 can be 50 giving LCM = 100

40 * 25, 12 numbers 80 and one number 40 giving LCM = 80

25 * 40 12 numbers can be 75 but 13th number is 100 LCM = 300

20 * 50  12 numbers 80 and one number 40 giving LCM = 80

So lowest LCM = 80 (12 numbers 80 and one number 40)