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user58865
user58865
00:19
Hi @mercio, very good point - thanks so much for explaining that. So, to avoid trivialities, $x$ is not a square root. Can it be some...rational number, though? Or, it must be an integer?
also, I am having some difficulty with the iterative / inductive process. Assuming x^2 = a is solvable modulo p, then it is true that x^2 = np + a. Here, I feel there is a simple iteration that I am overlooking...
...in order to show that x^2 = a is solvable modulo p^2...
let's call the solution (mod p) "k"
then I have that k^2 = mp+a, where m is an integer
mercio
mercio
00:40
usually, equations modulo things are about integers, (not rational number and even less irrational numbers)
your problem is that you are still trying to prove something that is not true
(what iterative/inductive process are you talking about)
 
2 hours later…
Erick Wong
Erick Wong
02:25
The claim is that if x in an integer such that x^2 is divisible by 3, then x is also divisible by 3, You will find that the example where x = \sqrt{3} is completely irrelevant to the truth or falsehood of this claim.

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