# Irrationality of the points on $x^2 + y^2 = 3$

On page 79 of his book "The Irrationals", Julian Harvil sets out to prove that all the points on the Cartesian plane of the circle described by $x^2 +y^2 =3$ are irrational... (paraphrased below)

1. Assume that a rational point ($\frac{p}{q},\frac{r}{s}$) lies on the circle.

2. Hence $p^2s^2 + q^2r^2 = 3q^2s^2$

3. Which we can restate as $a^2 + b^2 = 3c^2$ where $a, b, c$ are all positive integers.

I follow this up to this point, but then he stipulates "since the sum of two even numbers and the sum of two odd numbers are each even, it must be that one of $a^2$ and $b^2$ is even and the other odd".

Why is it necessary for the sum of $a^2 + b^2$ to be odd?

Update My mistake - he stipulates that $a^2$ and $b^2$ may not both be even as well as may not both be odd - have correctly quoted him now

• Oh... oops. I was reading that as $x^2 + y^2$ instead of $a^2 + b^2$. So, if anyone saw my answer, please ignore it, as it's supremely wrong. :) Aug 3 '13 at 14:17

Note that if $a,b$ are both odd then $a^2+b^2\equiv2\bmod 4$,but since $3c^2\equiv 0\bmod 4$ if c is even and $3c^2\equiv 3\bmod 4$ if c is odd, this cannot occur.

And if $a,b$ are both even then $a,b,c$ cannot be coprime, which (the condition of coprime) is (usually) the start point of Fermat's method of descent (and also some other methods as well).

And indeed we don't need such property to prove that there is no integers solution of $a^2+b^2=3c^2$. Assume first there exists such $(a,b,c)$ and assume further they are coprime. Since $a^2+b^2\equiv0\bmod3$, we must have $a,b\equiv0\bmod3$, which in turn imply that $c\equiv0\bmod3$. And that is a contradiction since $(a,b,c)$ are coprime.

• Why do $p,q,r,s$ have to be coprime? Without further argument, $q$ and $s$ might both be even. Aug 3 '13 at 14:20
• @TonyK you are right, thanks. Aug 3 '13 at 14:29
• Your edit about Fermat's method of descent has me completely baffled! Aug 3 '13 at 14:34
• It all makes sense now! Apart from that second paragraph...Perhaps you should just delete it? Aug 3 '13 at 15:47
• @TonyK Perhaps my English is too poor to express what I mean. I mean that when we want to get a contradiction using Fermat's method of descent, usually we consider minimal solutions first, in which case $a,b$ cannot be both even, but not in general. Aug 3 '13 at 15:57

If $a$ and $b$ are odd, then since $a=ps$ and $b=qr$ it must be that all of $p,s,q$ and $r$ are odd. Thus $c=qs$ would also be odd.

There are three cases to consider for the form $a^2+b^2=3c^2$

If $a$ and $b$ are both even, then $c$ must be even, and we can divide the whole equation through by $4$. Note that this isn't barred the way the question is put, but let's look for a minimal solution. A minimal solution cannot have both $a$ and $b$ even.

If $a$ and $b$ are both odd, we consider the equation modulo 4. The left-hand side is equivalent to $2$. The right-hand side is equivalent to $3$ ($c$ odd) or $0$ ($c$ even). No case of equality is possible.

We remain with the case of one odd and one even.

Note that odd squares are equivalent to $1$ modulo $8$ - and working mod $4$ or $8$ often gives a quick feasibility check on equations involving integer squares.

• Why is it necessary for the solution to be minimal? Aug 3 '13 at 14:31
• @adrianmcmenamin - it isn't, but this would be a standard reduction. If both $a$ and $b$ are even, there is also a solution where one of $a$ and $b$ is odd - keep dividing by $4$ until you can't any more - so if it is convenient we can look for that one to control the number of cases we are dealing with. Aug 3 '13 at 14:33
• I don't think it's a necessary stipulation, though. If $a$ and $b$ are even then RHS is 0 mod 4 while LHS is 2 mod 4. I think! Aug 3 '13 at 14:44