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are you asking the proof of

if p is a prime number then $(a_1+a_2+a_3+\cdots+a_n)^p\equiv (a_1^p+a_2^p+\cdots+a_n^p)\mod\ p$ ?

Here it is:

$(a_1+a_2+a_3+\cdots+a_n)^p=(a_1+b_1)^p$

where $b_1=a_2+a_3+\cdots+a_n$ $\equiv(a_1^p+b_1^p)\mod p$

$\equiv(a_1^p+b_1^p)\mod p$

$\equiv[a_1^p+(a_2+a_3+\cdots+\a_n)^p]\mod p$

$\equiv[a_1^p+(a_2+c_2)^p]\mod p$ where $c_2=a_3+a_4+\cdots+a_n$

$\equiv (a_1^p+a_2^p+c_2^p)\mod p$

continuing like this we get $(a_1+a_2+a_3+\cdots+a_n)^p\equiv (a_1^p+a_2^p+\cdots+a_n^p)\mod\ p$

Or if you just want the proof of $(a+b)^p\equiv (a^p+b^p)\mod p$

Let x=a² and y=b²

So now a+b²=7 and a²+b=11 so b=11-a²

And a+(11-a²)²=7

a^4 - 22a^2 + a +114=0

And a=3 is its solution. So x=9 and y=4

x^2 = x+x+x+x+................x this is in correct as x is not a constant. So as x changes the number of x also changes

"mathematically flawless"????

Who said to you that it is mathematical flawless???????

The flaw is that these rules are not for 1 or infinity.

first we don't have to use AM-GM inequality or the sign will be other way around.

The fallacy lies in ignoring the fact that the number of x's being added is not constant. Not only is x changing; the number of x's is also changing.

The fallacy lies in ignoring the fact that the number of x's being added is not constant. Not only is x changing; the number of x's is also changing.

well the subject is quadratic :P

please see the person's class/std before posting any replies.

not again..

the question is = 9x-3

And you have taken (4x^2+5x+1)^1/2 - 2(x^2-x+1)=9x-3

this should be (4x^2+5x+1)^1/2 - 2[(x^2-x+1)]^1/2=9x-3

Q2) This is a type of mordell's equation and it has only one solution for the constant k=2

(5,3) is the only solution for this

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