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
$(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$
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
We have
${2p}\choose{p}$$=\frac{(2p)!}{p!p!}$ bu definition
$=\frac{p!\prod_{k=p+1}^{2p}k}{p!p!}$ bu expansion
$=\frac{\prod_{k=p+1}^{2p}k}{p!}$ cancelling p! from both numerator and denominator
$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!}$ by taking 2 p ouut
working in mod p we get $(p+n) \equiv n \pmod p$
So we get
$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!} \pmod p$
$=\frac{2p\prod_{k=1}^{p-1}k}{p!} \pmod p$
$=2\frac{p\prod_{k=1}^{p-1}k}{p!} \pmod p$
$=2\frac{p!}{p!} \pmod p$
$=2$
Proved
When we break a number into smaller parts the product become larger as number of parts becomes larger and each is smaller however no part can be 1 as it shall provide a smaller value. so we can break it into 1012 parts and each is 2 giving a product $2^{1012}$.
However there is exception to ir as $2^3 \lt 3^2$ so we can group terms of 3 getting 674 3 and left with 1 2 giving a product $3^{674} * 2$
Proof:
A cube is either even or odd
If it is odd we can write it as $2n + 1$ which can be written as $(n+1)^2 - n^2$
If it is even the the number is even say $2n$ so cube is $8k$ where $n= k^3$ can be written as $(2k+1)^2 -(2k-1)^2$
We are given $\frac{x}{y+7} + \frac{y}{x+7}= 1$
or $x(x+7) + y(y+7) = (x+7)(y+7)$
or $x^2+7x + y^2 +7y = xy + 7x + 7y + 49$
or $x^2 + y^2 -xy = 49$
now $x^2- xy$ we need to form in form of squares by adding some teerm
let us multiply by 4 to above to get
$4x^2 + - 4xy + 4y^2 = 196$
or $4(x^2-4xy + y^2) +3y ^2 = 196$
or $(2x-y)^2 + 3y^2 = 196$
as it is sum of squares we need to check a finite number of values(fron pair of values we consider only positive ones)
$y = 3$ gives $2x-y =13$ giving $x = 8, y = 3$
$y= 5$ gives $2x-y = 11$ giving$ x = 8, y = 5$
$y = 7$ gives $2x -y = 7$ giving $x = y = 7$
$y = 8$ gives $2x -y = 2$ giving $x = 5 y = 8$
or $2x - y = -2$ giving $x = 3 and y = 8$
so the solutions are $(x=8,y=3), (x=8,y=5), (x=7,y=7),(x=5,y=8), (x=3,y=8)$
https://www.quora.com/What-is-the-equation-of-a-circle-which-passes-through-three-points-0-0-a-0-and-0-b
Because one point $(0,0)$ is origin, one point $(a,0)$ lies on x-axis and another $(0,b)$ on y axis it is clear that is is a right angled triangle. so the circumcenter is the midpoint of the hypotenuse that is $(\frac{a}{2},\frac{b}{2})$
so the equation of the circle is
$(x-\frac{a}{2})^2 +(y-\frac{b}{2})^2=(0-\frac{a}{2})^2 + (0-\frac{b}{2})^2 $ the RHS is square of distance of origin to to the centre
simplifying we get
$x^2-ax+y^2-by=0$
