Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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 $2$.
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!
$\endgroup$
Apr 20, 2018 at 21:04
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:
$${(k+1)^2\over{k^2}}=2\Leftrightarrow {{k+1}\over{k}}=\pm \sqrt2.$$
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
$$f(1)\gt f(2)=9/4\gt2\gt16/9=f(3)\gt f(4)\gt\cdots$$
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.
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.
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.
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.
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.
