somebody gave me this answer to the question in the title:
A2A: Observe that the remainder you get when you divide any square by 7 is never 3. You only need to observe it for the squares of the first 7 integers.
You also have to know that √m is rational if and only if m is a perfect square.
why does the remainder need to be 3 for it to be a perfect square?
In: 0
Assume sqrt(7n+3) is rational
Therefore sqrt(7n+3)=a/b where a and b are coprime integers (just the definition of a rational number)
7n+3 = a^2 /b^2
a^2 = b^2 (7n + 3)
Therefore 7n+3 | a^2 (7n+3 divides a^2 ) this is necessary because a is an integer, therefore a^2 must be one too, and we can follow the same logic for b and 7n+3. An integer times an integer results in an integer that is divisible by both integers and all of the factors of both integers.
Also, as a result, 7n+3 | a
Therefore a = k*(7n+3) where k is a positive integer
So 7n+3 = k^2 (7n+3)/b^2
7n+3 cancels, and 1 = k^2 /b^2
Therefore k|b
If k|b then b|a
If b|a, then a and b are not coprime, giving us a contradiction.
Therefore sqrt(7n+3) is irrational
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