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I am trying to prove the following:

Prove that there do not exist integers $x$, $y$ such that $x^2 = 5y^2$.

I have done the following up until now, but am unsure where to go from here:

By definition, it can be seen that $x^2$ is not even. Thus, $x^2$ must be odd. Suppose that $x$ is even. Then $x=2n$ for some $n \in Z$. Then, $x^2 = 4n^2$. So, $x^2$ is even. We have proved the contrapositive and therefore, $x$ is odd.

Suppose that there does exist a pair of integers $x,y$ such that $x^2 = 5y^2$. Choose a pair such that $x+y$ is minimal.

As $x$ is odd, $x=2t+1$ for some positive integer $t$.

But then $(2t+1)^2 = 5y^2$ so then $y^2 = \left (\frac{(2t+1)^2}{5}\right)$

Now I don't know how to use $y^2$ to actually prove that the theorem is true.

bawse
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  • see my response to http://math.stackexchange.com/questions/1310014/what-is-the-most-rigorous-proof-of-the-irrationality-of-the-square-root-of-3 – Mark Joshi Oct 24 '15 at 09:55
  • The number of prime divisors of $x^2$ is even whereas their number in $5y^2$ is odd. – Michael Hoppe Oct 24 '15 at 09:57
  • @MichaelHoppe I don't see how that can be used in this situation, could you explain a bit further please? – bawse Oct 24 '15 at 10:01
  • Small comment, there are integers $x$ and $y$ such that $x^2=5y^2$. But no non-zero integers. – André Nicolas Oct 24 '15 at 10:09
  • If the number of prime divisors of two natural numbers isn't equal the two numbers can't be equal, too – Michael Hoppe Oct 24 '15 at 10:09
  • @MichaelHoppe AH i see, but how can you tell if the number of prime divisors of a number are even or odd? – bawse Oct 24 '15 at 10:10
  • @AndréNicolas ah yes thank you, i should have made that more clear – bawse Oct 24 '15 at 10:12
  • Alternative: If $x^2= y^2+(2y)^2$ then $x$, $y$, and $2y$ would be a Pythagorean triple, hence $x/y$, $1$, and $2$ we be such a triple as well. – Michael Hoppe Oct 24 '15 at 10:12
  • Hmm, I would have much preferred to use the prime divisors explanation you mentioned a couple of comments ago – bawse Oct 24 '15 at 10:17

1 Answers1

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We give a proof that is perhaps along the lines you intended.

Suppose that there are positive integers $x,y$ such that $x^2=5y^2$. Then there are such integers with $x+y$ minimal.

Because $5$ divides $5y^2$, it divides $x^2$. Thus, because $5$ is prime, it divides $x$. Let $x=5x_1$. Then $25x_1^2=5y^2$, so $y^2=5x_1^2$.

This contradicts the minimality of $x+y$, because $y+x_1\lt x+y$.

André Nicolas
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