Proving Square root of 3 is irrational

Anyone who has ever taken high school mathematics at one point or other must have encountered the famous theorem of square root of 2 is irrational number. In fact this is considered to be one of the most famous and elegant proof in mathematics because of its simplicity. It also demonstrates the "method of contradiction", one of the most used method for proving, when all else doesn't seem to work.
The irrationality of square root of 3 can be proven similarly. The catch is when one does (p/q)^2 =3, we get p^2 = 3*q^2 which implies that either p^2 and q^2 are both even or both are odd. So we have two cases instead of one.
Case 1: Assuming both are even. But then if two numbers are even one can always divide them by 2 and so our initial assumption that p/q have no common factor is wrong. Which so implies that p and q cannot be both even.
Case 2: Now assuming both as odd. Thus p can be written as (2*m+1) and q can be written as (2*n+1) and our equation p^2= 3*q^2 can be written as
(2*m+1)^2 = 3 *(2*n+1)^2
4m^2 +4m+1 = 3(4n^2+4n+1)
4m^2+4m+1 = 12n^2+12n+3
4m^2+4m = 12n^2+12n+2
2(m^2+m) = (6n^2+6n+1)
Now left side is always even for all values of m, right side is always odd for all values of n, so this equality is never possible for any integer values of m and n and hence our assumption that p and q are odd is invalid.
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