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The goal: Prove that there is no integer $k$ such that ${(k+1)^2\over{k^2}}=2$.

My proof: If ${(k+1)^2\over{k^2}}=2$, then ${{k+1}\over{k}}=\sqrt2$, and if $k$ is an integer, $k+1$ is also an integer. This implies that $\sqrt2$ is a rational number, which is also provably false.

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    $\begingroup$ It works for me. $\endgroup$ – saulspatz Apr 19 '18 at 17:17
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    $\begingroup$ More directly: $\frac {(k+1)^2}{k^2}=2\implies k^2+2k+1=2k^2\implies k^2-2k-1=0$ and it is easy to see that this has no integral roots. $\endgroup$ – lulu Apr 19 '18 at 17:18
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    $\begingroup$ Nitpick (but a valid one): "Given some integer k, prove that there is no k such that" is garbled, and nonsensical and inconsistant. You meant to say simply "prove there is not integer $k$ so that...." $\endgroup$ – fleablood Apr 19 '18 at 17:24
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    $\begingroup$ nice! Good thought process and problem formulation. And yes, you proved it satisfactorily. Hmm, it'd be kind of a fun problem to trying to figure out when $\frac {(k+d)^2}{k^2} = m$. $m$ must be a perfect square. But what relation can $k$ and $d$ have? Can $d$ ever equal $1$? $\endgroup$ – fleablood Apr 19 '18 at 17:46
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    $\begingroup$ $(k+d)^2/k^2=m=x^2$, then $(k+d)/k=n$ and $k+d=kn$ and we require $d=k(n-1)$ i.e. to have the squares of two numbers separated by $d$ equal a perfect square, $d$ itself must be a multiple of the first number. This is only true for $d=1$ if $k=1$, giving $4/1=2^2$ in that case. AFAICT anyway... @fleablood $\endgroup$ – Nij Apr 20 '18 at 7:35
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How about solving the equation: $(k+1)^2=2k^2$ for $k\ne 0$?

$$(k+1)^2=2k^2$$

$$k^2+2k+1=2k^2$$

$$k^2-2k-1=0$$

$$k^2-2k+1=2$$

$$(k-1)^2-(\sqrt{2})^2=0$$

$$(k-1-\sqrt{2})(k-1+\sqrt{2})=0$$

Then $k=1+\sqrt{2}$ or $k=1-\sqrt{2}$, neither of them are integers, because you can prove that $\sqrt{2}$ is an irrational number (but knowing that it is not an integer is enough here).

For your method, the only minor mistake you made is:

$${(k+1)^2\over{k^2}}=2\Leftrightarrow {{k+1}\over{k}}=\pm \sqrt2.$$

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    $\begingroup$ My initial goal wasn't proving it false, the form my proof took was more a consequence of me trying to find a number that solved it. This is a lot better constructed than mine :). $\endgroup$ – user189728 Apr 19 '18 at 17:22
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    $\begingroup$ @user189728 Well, there are not many questions here I'm actually able to solve $\endgroup$ – Trần Thúc Minh Trí Apr 19 '18 at 17:26
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Another argument: Either $k$ or $k+1$ is even, the square of that is divisible by $4$, the other is odd, so the ratio can't be $2$.

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    $\begingroup$ That's probably the simplest proof possible, I love it! $\endgroup$ – user189728 Apr 19 '18 at 17:42
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    $\begingroup$ @user189728 Perhaps worth noting that it's the exact same argument used to prove that $\sqrt{2}$ is irrational. $\endgroup$ – Najib Idrissi Apr 19 '18 at 18:33
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    $\begingroup$ I found this argument a little hard to follow. To clarify: if k+1 is odd then the fraction is odd/even and that is never an integer, so never 2. If k+1 is even then the fraction has a factor of 4 on the top and an odd number on the bottom; in that case the fraction is either not an integer, so it is not 2, or it is an integer divisible by 4, in which case it is not 2. So either way, the ratio cannot be 2. Nice argument! $\endgroup$ – Eric Lippert Apr 20 '18 at 21:04
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Your proof is OK, but there is no need to invoke the irrationality of $\sqrt2$. It's enough to note that for

$$f(k)={(k+1)^2\over k^2}=\left(1+{1\over k}\right)^2$$

(with $k\not=0$), we have $f(k)\lt1$ if $k\lt0$ and

$$f(1)\gt f(2)=9/4\gt2\gt16/9=f(3)\gt f(4)\gt\cdots$$

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Just to be different:

$\frac {(k+1)^2}{k^2} = \frac {k^2 + 2k + 1}{k^2} = 1 + \frac {2k + 1}{k^2}$

So $\frac {(k+1)^2}{k^2} = 2\implies \frac {2k + 1}{k^2} = 1 \implies$

$2k + 1 =k^2$ and as $k \ne 0$ (otherwise $2 = 0$ which isn't true)

$2 + \frac 1k = k$. But if $k \ne \pm 1$ then $\frac 1k$ is not an integer.

And if $k = \pm 1$ then $\frac 1k = k$ and $2 = k -\frac 1k = 0$ which is impossible.

....

But seriously...

$\frac {(k+1)^2}{k^2} = (\frac {k+1}{k})^2 = 2$ implies there is a rational number whose square is $2$ which is well-known to be false, if a perfect and irrefutable proof. And probably the absolute easiest.

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    $\begingroup$ +1 for the last sentence. $\endgroup$ – John R. Strohm Apr 20 '18 at 22:36
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If you want to invoke the irrationality of $\sqrt 2$, you can prove a much stronger result, that no two perfect squares $m^2$ and $n^2$ can exist so that $\frac{m^2}{n^2} = 2$. It's as simple as taking the square root of both sides and noting the rationality of one side and the irrationality of the other.

The reason I am bringing this up is because you mentioned in a comment that this was a problem you came up with, so I thought it might be nice to lead you to an even stronger result than the one you conjectured.

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    $\begingroup$ So an even stronger result is that if the ratio of 2 perfect squares is whole, it must itself be a perfect square, which can be proven by looking at the prime decomposition. $\endgroup$ – yoniLavi Apr 20 '18 at 20:42
  • $\begingroup$ Henrik's answer is a really good answer and with that answer there, this answer is not needed. It's harder to prove the more general result than the result the question asked you to prove. Maybe if you stated that it was already known that $\sqrt 2$ is irrational then used it to prove the result you were asked to prove, the answer would be a bit better for those people who already knew that $\sqrt 2$ was irrational and then gimusi's answer wouldn't add anything to your answer and maybe be worthy of deletion. $\endgroup$ – Timothy Apr 21 '18 at 3:03
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And yet another way... if $k\ge3$ then $$2\le\frac23k\ ,\qquad 1\le\frac19k^2$$ so $$(k+1)^2=k^2+2k+1\le k^2+\frac23k^2+\frac19k^2=\frac{16}9k^2<2k^2\ .$$ This leaves $k=1,2$ as the only options, and it's easy to check that they don't work either.

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  • $\begingroup$ You need to exclude negative integers first. $\endgroup$ – user21820 Apr 20 '18 at 6:59
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    $\begingroup$ @user21820 Indeedn, but that's easily done (t makes $\frac{(k+1)^2}{k^2}<1$). The point is that David's proof does not make use of the fact that $\sqrt 2$ is irrational. $\endgroup$ – Hagen von Eitzen Apr 22 '18 at 4:33
  • $\begingroup$ @HagenvonEitzen: I know it is easily done. But for completeness, it should have been mentioned even if not proven. That's all I meant. $\endgroup$ – user21820 Apr 22 '18 at 16:31
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Yes it is absolutely correct by the irrationality of $\sqrt 2$.

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    $\begingroup$ Thanks! I added in the link to Wikipedia to the OP $\endgroup$ – user189728 Apr 19 '18 at 17:21
  • $\begingroup$ That's fine, well done! $\endgroup$ – gimusi Apr 19 '18 at 17:22
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While the other answers are focused on how this can be proven easier I am focusing on the question if your proof is correct.

Your proof contains an error:

If ${(k+1)^2\over{k^2}}=2$, then ${{k+1}\over{k}}=\sqrt2$

However ${{k+1}\over{k}}$ may also be $(-\sqrt2)$ in this case.

If you however argue that neither $\sqrt2$ nor $(-\sqrt2)$ are rational numbers your proof will work.

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No sequential integer squares will ever have a ratio of two. In fact, no integer squares will ever have a ratio of two. That is to say, no integer square will ever be exactly twice the size of another.

Start by assuming $a^2=2b^2$. The quantity $b^2$ has the prime factorization of $list*list$, where "list" is the prime factorization of $b$, multiplied in long form. If $b$ were 10, for example, the element "list" would be replaced with $5*2$.

Because there are two of each prime factor from $b$, we deduce that $b^2$ has an even number of prime factors. The same argument may show that $a^2$ will also have an even number of prime factors.

Obviously, $a^2$ cannot have the prime factorization $primelist*primelist*2$, as this is an odd number of prime factors. Contradiction.

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