2024/055) if $2n+ 1$ is a perferct square show that $n+1$ is sum of 2 squares when n is an integer

 We know $2n+1$ is odd so it is square of an odd number say $(2k+1)^2$

So $2n+1 = (2k+1)^2 = 4k^2 + 4k +1$

or $2n= 4k^2+ 4k$

or $n = 2k^2 + 2k$

or $n + 1=  2k^2+2k + 1 = k^2+(k^2+2k+1) = k^2+(k+1)^2$

Proved 

2024/054) Find x and y such that $LCM(x,y) = LCM(12x,5y) = 720$

 As $LCM(x,y) = 720$ we need to find prime factors of 720

$720=2^4 * 3^2 *5$

So $x=2^a3^b5^c\cdots(1)$

And $y=2^d3^e4^f\cdots(2)$

Where $max(a,d) = 4\cdots(3)$

  $max(b,e) = 2\cdots(4)$

 $max(c,f) = 1\cdots(5)$'

Now

$LCM(12x,5y) = LCM(2^{a+2}3^{b+1}5^c $$, 2^d3^e5^{f+1})= 2^4 * 3^2 *5$

So $max(a+2,d) = 4\cdots(6)$

$max(b+1,e) = 2\cdots(7)$

$max(c,f+1) = 1\cdots(8)$

From (3) and (6) we get $d=4$ and $a \le 2$

From (4) and (7) we get $e=2$ and $b\le 2$

from (5) nd (8) we get $c=1$ and $f=0$

So we get following pairs $(2^0 * 3^0 * 5, 2^4*3^2)$ that is $(5,144)$

$(2^1 * 3^0 * 5, 2^4*3^2)$ that is $(10,144)$

$(2^2 * 3^0 * 5, 2^4*3^2)$ that is $(20,144)$

$(2^0 * 3^1 * 5, 2^4*3^2)$ that is $(15,144)$

$(2^1 * 3^1 * 5, 2^4*3^2)$ that is $(30,144)$

$(2^2 * 3^1 * 5, 2^4*3^2)$ that is $(60,144)$

 

    

 

2024/053) Given $ab + bc + ca = abc$ find the value of $\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$

 We are given

$ab+bc+ca = abc$

Hence $bc(a-1) = abc - bc = ab + ac = a(b+c)$

Hence

 $\frac{(b+c)}{bc(a-1)} =  \frac{(b+c)}{a(b+c)} = \frac{1}{a}$

or 

 $\frac{(b+c)}{bc(a-1)} = \frac{1}{a}\cdots(1)$

 similarly

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

and

 $\frac{(a+b)}{ab(c-1)} = \frac{1}{c}\cdots(3)$

Adding (1),(2),(3) we get

 $\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$

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

$=\frac{bc+ca+ab}{abc} = \frac{abc}{abc}=1$

2024/052) Solve in integers $x^3+3367=2^n$

We are given $x^3+3367=2^n$

We know $3367=7 * 13 &*37

So let is work $x^3=2^n \pmod 7$

Working in mod 7 we have $x^3 \in \{1,-1,0\}$ and  $2^n \in \{1,2,4\}$ so we get 1 as common and for that n has to be multiple of 3.

So we get

$x^3= 2^{3k} \pmod 7$

Going back to the original equation we get

$x^3 + 3367 = 2^{3k}$

Or $(2^k)^3 - x^3 = 3367$

Or $y^3 -x^3 = 3367$ where $y = 2^k$

As $(y-x) | y^3-x^3$ so $y-x |  3367\cdots(1)$

Further 

As we know $(y-x)^3 = y^3-3y^2x + 3yx^2 - x^3 = (y^3-x^3) - 3yx(y-x) \lt y^3-x^3$ when $y-x \gt 0$

So $y-x\lt 15\cdots(2)$

So $y-x \in \{1,7\}\cdots(2)$ from (1) and (2)  

 $y-x =1$

gives $x^3 + 3367 = (x+1)^3 = x^3 + 3x^2 + 3x + 1$

Or $3367 = 3x^2+ 3x + 1$

Or $3x^2+3x = 3366$

Or $x^2 + x = 1122$

x = 33 and y = 34 and y is not power of 2 so not a solution

$y-x=7$

gives

$x^3 + 3367 = x^3 + 21x^2 + 147 x + 343$

or $3024 = 21x^2+ 147x

or $x^2+ 7x = 144 $

or $x = 9$ or $x=-16$

only positive value admissible giving x = 9 and y = 16

$2^n = y^3$ giving $2^n = 2^{12}$ or n = 12

Hence x=3 , n = 12