The discussion centers on proving that if \( \frac{x^2+y^2}{1+xy} \) is an integer for positive integers \( x \) and \( y \), then it must be a perfect square. Participants debate the completeness of Albert's proof and explore various approaches to the problem, including polynomial representations. An example of the pair \( (8, 30) \) is presented, demonstrating that it yields a perfect square, yet does not conform to the expected form of \( (u, u^3) \). The conversation also touches on the existence of chains of solutions and the conjecture that all solutions may follow a specific recursive pattern. The complexity of the problem is acknowledged, with participants expressing interest in finding a concise proof.
#1
anemone
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If $x,\,y$ are positive integers such that $\dfrac{x^2+y^2}{1+xy}$ is an integer, then $\dfrac{x^2+y^2}{1+xy}$ is a perfect square.
If $x,\,y$ are positive integers such that $\dfrac{x^2+y^2}{1+xy}$ is an integer, then $\dfrac{x^2+y^2}{1+xy}--(1)$ is a perfect square.
let :$x=\left | k^3 \right |,\,\ y=\left | k\right |$
or: $x=\left | k \right |,\, y=\left | k^3 \right |$
then :$(1)\,\,becomes :\,\, k^2$
where $ k \in Z$
and $k\neq 0$
#3
anemone
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Thanks for participating, Albert!
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.:)
#4
Albert1
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anemone said:
Thanks for participating, Albert!
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
#5
anemone
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Ah, I see what you meant now, thanks for the response. And thanks again to participating in my challenge.
#6
kaliprasad
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Albert said:
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
#7
anemone
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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!
#8
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anemone said:
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
#9
Opalg
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anemone said:
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.
#10
Albert1
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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 $y=x^3$
#11
Nono713
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Albert said:
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 :)
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#12
Nono713
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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.
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#13
anemone
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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.