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I get $|x+yi|=|z-u\omega|$. Can we solve the problem from this?

orl wrote: Consider the system $ x + y = z + u,$ $ 2xy \& = zu.$ Find the greatest value of the real constant $ m$ such that $ m \leq x/y$ for any positive integer solution $ (x,y,z,u)$ of the system, with $ x \geq y$.

i have another solution... note that $ x,y$ are the solutions to the equation $ a^2 - a(z + u) + \frac {zu}{2} = 0$.. calculating its discriminant it follows that it equals $ \Delta = z^2 + u^2$, so $ (z,u) = (m^2 - n^2,2mn)$ or $ (2mn,m^2 - n^2)$ for som positive integers $ m,n$, $ m > n$... in any of the cases $ x = \dfrac{z + u + \sqrt\Delta}{2} = m^2 + mn$ and $ n = \dfrac{x + u - \sqrt\Delta}{2} = mn - n^2$... it follows that $ q = \dfrac{x}{y} = \dfrac{\left(\frac {m}{n}\right)^2 + \left(\frac {m}{n}\right)}{\left(\frac {m}{n}\right) - 1}$... note that $ q\geq (1 + \sqrt2)^2$ is equivalent to $ \left(\frac {m}{n} - (1 + \sqrt2)\right)^2\geq 0$ which is true... if $ m = (1 + \sqrt2)^2 + \epsilon$ for some $ \epsilon > 0$ we would have that $ \dfrac{\left(\dfrac{m}{n} - (1 + \sqrt2)\right)^2}{\frac {m}{n} - 1}\geq \epsilon$... however we can take a sequence of rationals that converges to $ 1 + \sqrt2$ and thereferore the last inequality is not satisfied by every rational number $ \frac {m}{n}$ and we're done...

Does anyone find two numbers $ m,n$ for which $ \frac{x}{y}<5.9$?

Official Solution: Squaring the first equation and then subtracting four times the second, we obtain $x^2-6xy + y^2 = (z-u)^2$, from which $(x/y)^2-6(x/y)+1=((z-u)/y)^2$. The quadratic $w^2-6w+1$ takes the value $0$ for $3\pm2\sqrt2$, and is positive for $w>3+2\sqrt2$. Because $x,y\geq1$ and the right side of $(*)$ is a square, the left side of $(*)$ is positive, and we must have $x/y>3+2\sqrt2$. We now show that $x/y$ can be made as close to $3+2\sqrt2$ as we like, so the desired $m =3+2\sqrt2$. We prove this by showing that the term $((z-u)/y)^2$ in $(*)$ can be made as small as we like. To this end, we first find a way to generate solutions of the system. If $p$ is a prime divisor of $z$ and $u$, then $p$ is a divisor of both $x$ and $y$. Thus we may assume, without loss of generality, that $z$ and $u$ are relatively prime. If we square both sides of the first equation, then subtract twice the second equation we have $(x-y)^2 = z^2 + u^2$. Thus $(z,u,x-y)$ is a primitive Pythagorean triple, and we may assume that $u$ is even. Hence there are relatively prime positive integers $a$ and $b$, one of them even and the other odd, such that $z = a^2-b^2; u = 2ab$; and $x-y = a^2 + b^2$. Combining these equations with $x + y = z + u$, we find that $x = a^2 + ab$ and $y = ab-b^2$. Observe that $z-u = a^2-b^2 -2ab = (a-b)|^2 -2b^2$. When $z-u = 1$, we get the Pell equation $1 = (a-b)^2-2b^2$, which has solution $a-b = 3, b = 2$. By well known facts, this equation has infinitely many positive integer solutions $a-b$ and $b$, and both of these quantities can be made arbitrarily large. It follows that $y = ab-b^2$ can be made arbitrarily large. Hence the right side of $(*)$ can be made as small as we like, and the corresponding value of $x/y$ can be made as close to $3+2\sqrt2$ as we like.

Mewto55555 wrote: Then, $z=2ab$ , $u=a^2-b^2$ , $x-y=a^2+b^2$ $x+y=x-y+2y=a^2+b^2+2y=2ab+a^2-b^2 \implies y=ab-b^2$ Similarly, $x=a^2-ab$ It is a simple solution, but there is a mistake, we get $x=a^2+ab$ for having $x+y=u+z=a^2+2ab-b^2$ Now $ \frac{x}{y}=\frac{a^{2}+ab}{ab-b^{2}} $ is homogenous and so $b=1$ and $a^2+(3+2\sqrt{2} \ge 2(1+\sqrt{2})a$ of AM-GM. Hence the solution was true, with those mistakes.

I dont understand how can we get arbitrarily close to $ 3+2\sqrt2 $. Somebody can explain?

eraydin wrote: I dont understand how can we get arbitrarily close to $ 3+2\sqrt2 $. Somebody can explain? Take $a=[(1+\sqrt{2})b]$ where $[]$ is to take the greatest smaller integer (entier) For $b$ very high, it works very good, for $10^5$ we get $x=82426406871,y=14142135624$ for ex. and that is a good thing.

Please check my (bit different) solution

Here is a solution with the slope trick In fact we will go further and characterize all solutions $(x,y)$; notice that if we scale $x,y,z,u$ all by the same constant, both equations still hold, so assume that we are working in $\mathbb Q^+$ and scale so that $x+y = z+u$, and $2xy=zu$; let $x+y = z+u = 1$, and $2xy=zu = 2p$; then, we see that \[ 1-8p = a^2, 1-4p = b^2, \]for squares $a,b$ by considering the discriminants of $w^2-w+p$ and $w^2-w+2p$, which are squares of rational numbers as the quadratics have rational roots; so, $8p\leq 1$; this implies that \[ y \leq \frac{1-\sqrt{1-4p}}{2} \leq \frac{1}{4}(2-\sqrt 2), \]so \[ 4+2\sqrt 2 \leq \frac{1}{y} = 1+\frac{x}{y}, \]so $\frac{x}{y} \geq \boxed{3+2\sqrt 2}$. Now, in order to show that $\frac xy$ can get arbitrarily close, it suffices to show that $a$ can get arbitrarily close to $0$ in the equation $2b^2-a^2 = 1$, where we take $(a,b)$ over $\mathbb Q^2$. As long as this holds, we can get $(1-a^2) = 2(1-b^2)$, so setting $8p = 1-a^2$ works, and since $a\to 0$, $p \to \frac 18$ and $y\to \frac 14(2-\sqrt 2)$, which will imply the desired. Let $\mathcal C$ denote the locus of $\{(a,b)\colon 2b^2=1+a^2\}$, then $O = (1,1)\in \mathcal C$. Furthermore, for any other point $P$ on $\mathcal C$, notice that $P$ is rational if and only if $PO$ has rational slope! So, let this slope be $s$, and intersecting we have $1+a^2 = 2b^2$, $(1-b) = s(1-a)$; since $(1-a^2) = 2(1-b^2)$, this implies $(1+a) = 2s(1+b)$. But solving yields \[ a = \frac{2s^2-4s+1}{2s^2-1}, \]so taking $s$ to be an arbitrarily good approximation of $1+\frac{1}{\sqrt 2}$ is sufficient.

Here is a solution with the chain rule.

https://math.stackexchange.com/questions/779352/problem-heron-of-alexandria/783168#783168

Right... this bound seems so random, but somehow it works- anyways, here's my solution- First, notice that, as $x^2+y^2+2xy=z^2+u^2+2uz=>(x-y)^2=u^2+z^2$. As $\frac{(x-y)^2}2=\frac{z^2+u^2}2 =>x-y>\sqrt{2uz}$ (by RMS-GM) $=> x-y>2\sqrt{xy}=>x^2+y^2-6xy>0=>m^2-6m+1>0$ (here $m=\frac{x}{y}$)... hence $m>3+2\sqrt{2}$. After I got here, I tried to find some nicer bound, because such an arbitrary value couldn't be the bound, right... well, not rather unexpectedly, I didn't get anywhere, and only when I tried to bash out the problem using the fact that $(u,z,x-y)$ is a Pythagorean triplet did I realize that somehow this bound must be the correct one. After I realised that, it was just a matter of noticing that, as $x-y=m^2+n^2$ and $u,z=2mn,m^2-n^2$ (using Pythagorean triplet $(z,u,x-y)$ and $x+y=u+z$) $=> \frac{x}{y} =\frac{m(m+n)}{n(m-n)} =\frac{l^2+l}{l-1}$ where $l=\frac{m}{n}$- notice that we can take rational $\frac{m}{n}=l$ as close to $\sqrt{3+2 \sqrt{2}}=1+\sqrt{2}$ as we want from the right while still satisfying the conditions of the problem, we can bring $\frac{x}{y}$ as close to $3+2 \sqrt{2}$ as we want, and thus we're done.

The answer is $\boxed{3+2\sqrt{2}}.$ $\emph{Proof of Bound: }$ Let $z=x-d$ and $u=y+d.$ Then, the second equation yields $$2xy=xy+dx-dy-d^2$$$$\implies (d+x)(d-y)=2d^2.$$Write $x=da$ and $y=db$ for rational $a,b.$ We have $$(1+a)(1-b)=2\implies a=\frac{1+b}{1-b}.$$Hence, $$\frac{x}{y}=\frac{a}{b}=\frac{1+b}{b(1-b)}.$$Now it is easy to check that this expression has a minimum of $3+2\sqrt{2}$ on the interval $(0,1).$ $\emph{Proof of Maximality: }$ Fix a positive integer $k,$ and let $m=\lfloor(10^k(2+\sqrt{2})).$ Check that $$(d,x,y)=(10^km, m^2-10^km, m-2\cdot 10^{2k})$$satisfies the equation. Moreover, as $k$ becomes large, $\frac{x}{y}$ becomes arbitrarily close to $3+2\sqrt{2}.$ Thus, our bound is the best possible.

The answer is $3+2\sqrt{2}$. Note that $$xy = \frac{zu}{2} \le \frac{(\frac{z+u}{2})^2}{2} = \frac{(z+u)^2}{8} = \frac{(x+y)^2}{8}$$thus $$(x+y)^2 \ge 8xy \Rightarrow x^2 - 6xy + y^2 \ge 0 \Rightarrow \left(\frac{x}{y}\right)^2 - 6 \frac{x}{y} + 1 \ge 0.$$Solving for $\frac{x}{y}$ (noting that $\frac{x}{y}\ge 1$) gives us the desired bound, proving sufficiency. It suffices to prove necessity. WLOG let $u>z$ and scale down so that $z=1$. It suffices to prove that we can find rational triples $(x,y,u)$ so that the equation is satisfied and $u$ can get arbitrarily close to $1$. Note that $x$ and $y$ are roots of the quadratic $a^2 - (u+1)a + \frac{u}{2} = 0$. For $x$ and $y$ to be rational, we must have the discriminant $(u+1)^2-4\cdot \frac{u}{2} = u^2+1$ be the square of a rational number. Claim: We can find Pythagorean triples $(a,b,c)$ where $c>a,b$ and $\frac{a}{b}$ is arbitrarily close to $1$. Proof. Consider only primitive triples because others won't help. Using Euclid's formula we have $a = m^2-n^2, b=2mn$ for some integers $m,n$. Therefore we need to prove we can find $m,n$ so that $\frac{m^2-n^2}{2mn} = \frac{1}{2} \left(\frac{m}{n} - \frac{n}{m}\right)$ is arbitrarily close to $1$, which is equivalent (by solving) that $\frac{m}{n}$ is arbitrarily close to $1+\sqrt{2}$. It is clear that an arbitrarily close rational approximation to $1+\sqrt{2}$ can be found. $\blacksquare$ For a Pythagorean triple, scale it down so that the smallest number is $1$ and the second leg length is $u$. Thus there are valid values of $u$ sufficiently close to $1$, finishing the problem.

We call $x+y=z+u$ as $(1)$ and $2xy=zu$ as $(2)$ $(1)^2-4(2)$ leads to $x^2-6xy+y^2=(z-u)^2, (\frac{x}{y})^2-6 \frac{x}{y}+1=(\frac{z-u}{y})^2\geq 0$ by solving the quadratics we can find $\frac{x}{y}\geq 3+2\sqrt{2}$, now we need to prove that $\frac{z-u}{y}$ can approach $0$ Then, we do similar thing, $(1)^2-2(2)^2$ tells that $(x-y)^2=z^2+u^2$ which is a pythagorean triple. Where there must be an number is divisible by $2$. Let $u=2ab, z=a^2-b^2, (x-y)=a^2+b^2$. As $x+y=z+u=a^2+2ab-b^2$, we just let $x=a^2+ab, y=ab-b^2$ Firstly, if $z=u, a^2-2ab-b^2=0$, divide $b^2$ for each term, $(\frac{a}{b})^2-2\frac{a}{b}-1=0$, there are only irrational roots which doesn't satisfy the question. Then we check if $y$ can approach infinity We let $z-u=1, (a-b)^2-2b^2=1$ which is the general form of a pell equation. It's obvious that the elementary solution for the equation is $a-b=3, b=2$ so this equation can have infinity solutions so $ab-b^2$ can definitely approach infinity So the maximum $\frac{x}{y}=3+2\sqrt{2}$

We have $x^2+2xy+z^2=z^2+2zu+u^2$ so $x^2-6xy+y^2=(z-u)^2\ge 0$ and so if $r=\tfrac{x}{y}$ then $r^2-6r+1\ge 0$ which means $r>3+2\sqrt 2.$ We claim that we can get the ratio to be arbitrarily close to $3+2\sqrt 2.$ $~$ We have $(x-y)^2=z^2+u^2$ so let $z=a^2-b^2,u=2ab,x-y=a^2+b^2.$ Note that $x+y=a^2+2ab-b^2$ so $x=a^2+ab,y=ab-b^2.$ Let $a/b=s$ where $s>1.$ Then, we can make $s$ arbitrarily close to $1+\sqrt{2}$ which makes $\frac{x}{y}=\frac{s^2+s}{s-1}$ arbitrarily close to $3+2\sqrt2$ which is what we wanted.

The answer is $3+2\sqrt2$. Note that $(x-y)^2=u^2+z^2$. This means $x-y=k(a^2+b^2)$, $u=2kab$, and $z=k(a^2-b^2)$. We have that $x+y=u+z=k(a^2+2ab-b^2)$. This means $x=k(a^2+ab)$, and $y=k(ab-b^2)$. This means $a > b$. We have that $$ \frac{x}{y} = \frac{\left( \frac{a}{b} \right) ^2 +\frac{a}{b}}{\frac{a}{b}-1} = 3+\left( \frac{a}{b}-1 \right) + \frac{2}{\left( \frac{a}{b} -1 \right)} \ge 3+2\sqrt2,$$so we have that $3+2\sqrt2$ works. Taking $\frac{a}{b}$ to be arbitrarily close to $\sqrt{2}+1$, and $k=1$, we get a working quadruple and $\frac{x}{y}$ arbitrarily close to $3+2 \sqrt2$, so $3+ 2 \sqrt2$ is the optimal bound and we are done.

Taking the discriminant of the quadratic with roots $u,z$ we easily obtain $\frac{x}{y}\ge \boxed{3+2\sqrt{2}}$ so it's enough to show that we can pick $x,y$ so that $\frac{x}{y}$ becomes arbitrarily close to this value. To show this, we show that we can pick arbitrarily large $x,y$ such that $x^2-6xy+y^2=1$. Note that $(3y+k,y)$ is always a solution where $k^2=8y^2+1$, and since there are infinitely many solutions to this we are done. $\blacksquare$

We have $x^2-6xy+y^2=(x+y)^2-8xy=(z+u)^2-4zu=(z-u)^2\ge 0$. Then if v=x/y we have $LHS/y^2=v^2-6v+1\ge 0$, or that v>3+2$\sqrt{2}$. Now it remains to find such a v. Note that $(a^2+b^2)^2=(x-y)^2=(x+y)^2-4xy=(z+u)^2-2zu=z^2+u^2=(a^2-b^2)^2+(2ab)^2$, so $(x-y,z,u)=((a^2+b^2),(a^2-b^2),(2ab))$. We also have that $a^2+b^2+2y=x-y+2y=x+y=z+u=a^2-b^2+2ab$, so $(x,y)=(a^2+ab,ab-b^2)$. If a/b=w, we can make w infinitely close to $1+\sqrt{2}$ which would then satisfy $v=\frac{x}{y}=\frac{a^2+ab}{ab-b^2}=\frac{a^2/b^2+a/b}{a/b-1}=\lim_{w->1+\sqrt{2}}\frac{w^2+w}{w-1}=\boxed{2\sqrt{2}+3.} \blacksquare$

The answer is $3+2\sqrt{2}$. Suppose that $z+u=4s$. Then, $$2xy=zu\leq 4s^2,$$so $$xy\leq 2s^2.$$Since $x+y=4s$, given that $x\geq y$, the minimum possible $x/y$ occurs when $x=s(2+\sqrt{2})$ and $y=s(2-\sqrt{2}),$ as we need to spread at least this far to get the product at most $2s^2$, so thus $m=3+2\sqrt{2}$ works since $x/y\geq 3+2\sqrt{2}.$ Now, consider $z=2s+2,u=2s-2$. Then, we need $$xy=2s^2-2,x+y=4s.$$Using Vieta and then solving the quadratic, we get $$x,y=2s\pm \sqrt{2s^2+2},$$and by Pell on $k^2-2s^2=2$, there exist arbitrarily large $s$ for which $x,y$ are both integers. Furthermore, $$\frac{x}{y}=\frac{2s+ \sqrt{2s^2+2}}{2s- \sqrt{2s^2+2}}=\frac{6s^2+4s\sqrt{2s^2}+2}{2s^2-2}.$$As $s$ gets large, this approaches $3+2\sqrt{2}$, hence done.

I claim the answer is $3 + 2\sqrt{2}$. Proof of sufficiency: Let $x + y = 2a$, in order to minimize $\frac xy$ we need to maximize $xy$, so we need to maximize $zu$, which occurs when $z = u = a$, so setting $x(2a - x) = \frac{a^2}{2}$ gives $x^2 -2ax + \frac{a^2}{2}$, quadratic formula gives $x = a \pm a\frac{\sqrt{2}}{2}$, taking the greater solution for $x$ yields $\frac xy = 3 + 2\sqrt{2}$ as the minimal value. Proof of necessity: We consider solutions over rationals instead, as all rational solutions correspond to integer solutions with the same value of $\frac xy$ by multiplying by a common factor. By quadratics, we explicitly determine $\frac xy$ in terms of $zu$ (it is minimized with increasing $zu$), also given fixed sum $z + u$ we know $zu$ is maximized when $z,u$ are closest together. So fix $z + u = 2$, we show there exist rational solutions with $z,u$ being arbitrarily close, and this finishes. We desire $\frac 12 z(2 -z) = x(2 - x)$, so for a rational solution we require $\sqrt{4 +2z^2 - 4z}$ to be a rational. Letting $z = \frac pq$, we desire $\sqrt{2p^2 - 4pq + 4q^2} = \sqrt{p^2 + (p - 2q)^2 }$ to be an integer. This has infinitely many solutions of the form $q = p - 1$ by Generalized Pell equation stuff, so we are done (since we can get $z$ as close to $1$ as desired).

Let $s = x+y$, $p = xy$, $A = x-y$, and $B = z-u$. We should have \begin{align*} s^2 - 4p &= A^2, \\ s^2 - 8p &= B^2, \\ \implies s^2 + B^2 &= 2A^2. \end{align*}In particular, any solution to the first two equations yields a solution $(x, y, z, u)$, since $\{x, y\}$ and $\{z, u\}$ are the sets of roots of $t^2 - st + p = 0$ and $t^2 - st + 2p = 0$ respectively. Furthermore, any solution to the third equation has $\nu_2(s) = \nu_2(A) = \nu_2(B)$ by infinite descent, so it does lead to an integer $p$ and therefore a solution. As such, we only need to analyze \[ s^2 + B^2 = 2A^2. \]This clearly implies $\tfrac{s}{A} < \sqrt2$, and Pell's equation promises we can get arbitrarily close to $\sqrt2$ while setting $B = 1$. Since \[ \frac xy = \frac{s+A}{s-A} = \frac{\tfrac sA + 1}{\tfrac sA - 1} \]is a decreasing function of $\tfrac sA > 1$, we always have \[ \frac xy > \frac{\sqrt2+1}{\sqrt2-1} = 3+2\sqrt2. \]Furthermore, we can get arbitrarily close via Pell, so $m = 3+2\sqrt2$ is optimal. $\blacksquare$