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

 

 

2024/051) Find integer solutions to $1+2^x+ 2^{2x+1} = y^2$

 We have 

 $1+2^x+ 2^{2x+1} = y^2$

Or  $2^x+ 2^{2x+1} = y^2-1$

Let us check one case x = 0 for which $2^x$ is odd that is 1 and we get $1 + 1+ 2 = 4 = y^2$ or 

(0, 2), (0, -2) are 2 solutions.

 $2^x(1+2^{x+1}) = y^2-1 = (y+1)(y-1)\cdots(1)$ in this x is positive

As LHS is even  number so RHS is even.

Now y+1 and y-1 both must be even and one has to be an odd multiple of power of 2 and other one odd multiple of 2.

So $y = 2^{x-1}p\pm 1$  where p is odd.

Or  $y = 2^{x-1}p +q$ where p is odd and q is $\pm 1$

Putting in the equation (1) we get

$2^x(1+2^{x+1}) = y^2-1 = (2^{x-1}p +q)^2-1 = 2^{2x-2}p^2 + 2^x pq$

Or $1+2^{x+1} = 2^{x-2} p^2 + pq$

Or  $1-pq = 2^{x-2}(p^2-8)$

q =1 gives $1-p = 2^{x-1}(p^2-8)=>(p^2-8)<0$ or $p=1$ which does not satisfy the condtion.

If q = -1 then we have

$1+p = 2^{x-2}(p^2-8)$ giving $1 + p > p^2 -8$

Or    $p^2-p -7 <=1$ this has $p  <= 3$ as p = 3 giving x = 4 and $y = \pm 23$

So 4 solutions  are $(0,2),(0,-2),(4,23), (4,-23)$

  

  

2024/050) If a+b+c =1 for positive numbers a,b c prove that $(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$

We have 

$a+b+c = 1$

So $1+a + b+c =  2$

Or $1+a = 2 - b-c = (1-b)+(1-c)$

because (1-b) and (1-c) are positive using AM GM inequality we have

$(1-b)(a-c) >= 2\sqrt{(1-b)(1-c)}$

or $(1+a) >= 2\sqrt{(1-b)(1-c)}\cdots(1)$

similarly 

$(1+b) >= 2\sqrt{(1-c)(1-a)}\cdots(2)$

and

$(1+c) >= 2\sqrt{(1-a)(1-b)}\cdots(3)$

Multiply above 3 equations we get

 $(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$

2024/049) Prove that there are infinite prime numbers of the form 4n+3

Let there be finite number of prime numbers of the form 4n+ 3 the number of numbers by k with $p_k = 4 q_k +3$

Now consider the number $4p_1p_2\cdots p_k -1 $

The number above is of the form 4n+3.

If the number is prime then we have found a prime number above the largest prime and we are done.

If it is not a prime number it has got some prime factors. 

All the prime factors cannot be of the 4n+1 because in that case product shall be of the from 4n+ 1

So it has got a prime factor of the form 4n+3 above $p_k$. which is again a conradiction.

Hence there are infinite prime numbers of the form 4n+3