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(Apologies for a further question on the same problem)

On page 79 of Julian Harvil's book "The Irrationals" he sets out to prove (by contradiction) that all the points on the circle described by $x^2+y^2=3$ are irrational.

To paraphrase his proof:

  1. Let $\left(\frac{p}{q},\frac{r}{s}\right)$ be a point on the circle, where $p,q,r,s$ are all integers

  2. Hence $(ps)^2 + (qr)^2 = 3(qs)^2$

  3. We can restate this as $a^2 + b^2 = 3c^2$ and we know that one of $a,b$ must be odd and the other even

  4. So $a^2 + b^2 = 4(m^2 + n^2 + n) + 1 = 4N + 1 = 3c^2$

  5. We can say that $c$ could be of the form $4M, 4M + 1, 4M + 2$ or $4M + 3$

    And this is where I get lost (this time):

  6. This means $c^2$ is of the form $4N, 4N + 1, 4N, 4N + 1$ so $3c^2$ must be of the form $4N, 4N + 3, 4N, 4N + 3$

Could someone explain the reasoning for this last line?

KReiser
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adrianmcmenamin
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1 Answers1

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Remember that the square $c^2$ of an integer $c$ is either $\equiv 0\pmod 4$ (if $c$ is even) or $\equiv 1\pmod 8$ (if $c$ is odd); in fact $1\pmod 4$ instead of $1\pmod 8$ is good enough here, as we obtain $3c^2\equiv 0\pmod 4$ or $3c^3\equiv 3\pmod 4$, but definitelys not $3c^3\equiv 1\pmod 4$.

Hagen von Eitzen
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