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
Let’s break it down, in order for √m to be rational, m must be a perfect square.
In this case m = 7n + 3, so 7n + 3 must be a perfect square.
One way we can write that is:
7n + 3 = PS
where n is an integer and PS is a perfect square. Let’s divide both by seven:
n + 3/7 = PS/7
That is, when we take our perfect square and divide it by seven, we’ll get an integer, n, and 3/7. Well that’s the same thing as saying a remainder of 3 after dividing by 7, for example, 5 divided by 3 is 1 remainder two which we could write as:
1 + 2/3 = 5/3.
So let’s look at the first seven integers, their squares, and the remainders when dividing by seven:
1 -> 1^(2) = 1 -> 1/7 = 0 + 1/7 (remainder 1)
2 -> 2^(2) = 4 -> 4/7 = 0 + 4/7 (remainder 4)
3 -> 3^(2) = 9 -> 9/7 = 1 + 2/7 (remainder 2)
4 -> 4^(2) = 16 -> 16/7 = 2 + 2/7 (remainder 2)
5 -> 5^(2) = 25 -> 25/7 = 3 + 4/7 (remainder 4)
6 -> 6^(2) = 36 -> 36/7 = 5 + 1/7 (remainder 1)
7 -> 7^(2) = 49 -> 49/7 = 7 + 0/7 (remainder 0)
So none of the remainders are 3, meaning you can’t write any of them in the form n + 3/7 = PS/7 where PS is a perfect square and n is an integer, which means you can’t write any of them in the form 7n + 3 = PS.
So why do we only have to check the first 7 integers? Well that’s because after that, the pattern of remainders (1, 4, 2, 2, 4, 1, 0) repeats.
A number can only be a perfect square with a remainder of 3 when divided by 7 if it is both A) a perfect square and B) a number that has a remainder of 3 when divided by 7
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
Remainder does not need to be 3 for it to be a perfect square – the opposite, actually: perfect squares **never** have remainder 3 (when divided by 7).
It’s just that the number `7n + 3` *does* have a remainder 3 (`7n` has remainder 0, `3` has remainder 3, so `7n + 3` has remainder 0+3).
But, because perfect squares never have remainder 3 (div. 7), `7n+3` is not a perfect square.
But only perfect squares have rational roots, so √(7n+3) is not rational.
Therefore, it is irrational.
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