But I don't see how we can relate $x$ and $y$ in $k$ in such a manner. Do you mind to explain it for me? Thanks.:)
let :$\dfrac {x^2+y^2}{1+xy}=k^2$
we have:$x^2+y^2=k^2(1+xy)$
$=x^2(1+\dfrac {y^2}{x^2})$
let $x^2=k^2$ then $\dfrac {y^2}{x^2}=xy$
or $y=x^3$
for $x,y \in N$
we get :$x=\left | k \right | $
$y=\left | k^3 \right |$
because of symmetry $x ,\,and,\, y$ are interchangeable
let :$\dfrac {x^2+y^2}{1+xy}=k^2$
we have:$x^2+y^2=k^2(1+xy)$
$=x^2(1+\dfrac {y^2}{x^2})$
let $x^2=k^2$ then $\dfrac {y^2}{x^2}=xy$
or $y=x^3$
for $x,y \in N$
we get :$x=\left | k \right | $
$y=\left | k^3 \right |$
because of symmetry $x ,\,and,\, y$ are interchangeable
by choosing the expression to be $x^2$ we are not considering all possible cases hence the given solution is wrong
Hmm...while I do have a solution that approached the problem quite different than Albert's, I think what Albert did is justifiable. But I could be wrong.
Do you have something, an example in mind to say that Albert has not included all possible cases in this problem? :)
And welcome back, kaliprasad! I was, initially, afraid you have gotten old of seeing my challenges and your return to MHB makes me very happy!
Hmm...while I do have a solution that approached the problem quite different than Albert's, I think what Albert did is justifiable. But I could be wrong.
Do you have something, an example in mind to say that Albert has not included all possible cases in this problem? :)
And welcome back, kaliprasad! I was, initially, afraid you have gotten old of seeing my challenges and your return to MHB makes me very happy!
I have not claimed that other value is possible but without proving that other value is not possible we cannot assume so
Hmm...while I do have a solution that approached the problem quite different than Albert's, I think what Albert did is justifiable. But I could be wrong.
Albert has shown that every perfect square can occur as a value of $$\frac{x^2+y^2}{1+xy}.$$ That is a neat result. But it does not answer the given problem, which is to show that perfect squares are the only natural numbers that can arise in this way. I have tried hard, without any success, to prove that result, which is in a sense the converse of Albert's result. I should very much like to see a solution of it.
if $\dfrac {x^2+y^2}{1+xy}=m\in N$
we have $x^2-mxy+y^2-m=0---(*)$
consider it as a polynomial respect to x with degree 2
then $x=\dfrac {my\pm \sqrt {m^2y^2-4(y^2-m)}}{2}$
for $x,y\in N$
we obtain $m=y^2$ and $x=my=y^3$
if we took (*) as a polynomial respect to y with degree 2 we have $m=x^2$ and $y=x^3$
if $\dfrac {x^2+y^2}{1+xy}=m\in N$
we have $x^2-mxy+y^2-m=0---(*)$
consider it as a polynomial respect to x with degree 2
then $x=\dfrac {my\pm \sqrt {m^2y^2-4(y^2-m)}}{2}$
for $x,y\in N$
we obtain $m=y^2$ and $x=my=y^3$
if we took (*) as a polynomial respect to y with degree 2 we have $m=x^2$ and $y=x^3$
What about the solution $(8, 30)$? In that case we have:
As you can see, most of the pairs here are of the form $(u, u^3)$, but there are four "rogue" pairs that are not of that form. I observe that each positive integer $u$ spawns a unique chain of solutions. And, in fact, starting from any integer $u$ and its associated trivial solution $(u, u^3)$, those chains of solutions obey a simple recurrence of the form:
This, of course, is a strict superset of the solutions given by Albert, and seems to be the dominant structure of the set of solutions. I conjecture that all solutions are of this form (exist in a chain for some $u \in \mathbb{N}$), and will try to find a proof of this. Perhaps an inductive proof might work, after showing that all chains are infinite and disjoint, if they are. I feel that this problem is actually pretty hard, but should we all fail to solve the challenge I look forward to anemone's solution which will without doubt be short and elegant :)
Got it, I think. There are probably a few aspects of the proof that could be clarified or are a bit handwavy, but I think the approach is understandable enough.
First assume $m < n$, without loss of generality, as the only solution of the form $(m, m)$ is $(1, 1)$ since if $1 < m < n$ then $mn < n^2$ hence $m^2 + n^2 \ne 1 + mn$. The other solutions for $m > n$ can be obtained by symmetry. Therefore, suppose we have a solution $(m, n)$ with $1 < m < n$ such that:
Then it is easy to observe that $(rm - n, m)$ is also a solution, as:
$$(rm - n)^2 + m^2 = r^2 m^2 - 2 r m n + n^2 + m^2 = r(r m^2 - m n + 1) = r((r m - n)m + 1)$$
And $(r m - n)m + 1 \ne 0$ since $r > 1$ and so:
$$\frac{(rm - n)^2 + m^2}{1 + (rm - n)m} = r$$
Thus if we apply this formula repeatedly we can generate a sequence of solutions for which the fraction also equals $r$. We must show that for every solution $(m_k, n_k)$ in the sequence, the property that $0 \leq m_k < n_k$ is maintained. We also prove this by induction, by posing:
By induction, this holds for every term of the series. Furthermore, the lower bound for $rm - n$ tends to zero as $m$ and $n$ get smaller and thus closer to each other, such that $m_k \geq 0$ in the limit. We have thus shown that the sequence is strictly decreasing, and that we will therefore eventually arrive at a solution of the form $(0, s)$ for some integer $s > 0$. As we have proven, the fraction for this term must equal $r$ too, so:
$$\frac{0^2 + s^2}{1 + 0 \cdot s} = r$$
In other words, $s^2 = r$, and so $r$ was a square.
Interesting and thorough proof,(Yes) thanks Bacterius!
Now it is my turn to show another proof (that I saw somewhere online) here:
Suppose that $\dfrac{x^2+y^2}{1+xy}=k$ is a counterexample of an integer which is not a perfect square, with max $(x,\,y)$ as small as possible. We may assume WLOG that $x<y$ for if $x=y$, then
$0<\dfrac{2x^2}{1+x^2}=k<2$ which forces $k=1$, a perfect square.
Now, $x^2+y^2-k(1+xy)=0$ is a quadratic in $y$, $y^2-kxy+x^2-k=0$. Let $y_1,\,y$ be its roots, so $y_1+y=kx$ and $y_1y=x^2-k$.
As $x,\,k$ are positive integers, suppose $y_1<0$ is incompatible with $x^2+y_1^2=k(1+xy_1)$. As $k$ is not a perfect square, suppose $y_1=0$ is incompatible with $x^2+0=k(1+0\cdot x)$.
Also, $y_1=\dfrac{x^2-k}{y}<\dfrac{y^2-k}{y}<y$
Thus, we have found another positive integer $y_1$ for which $\dfrac{x^2+y_1^2}{1+xy_1}=k$ and which is smaller than the smallest max $(x,\,y)$. This is a contradiction.
It must be the case, then, that $k$ is a perfect square.
But I could be wrong.