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The problem is to prove non-existence or to show that there exists two distinct pairs(up to permutation) of natural numbers $(a, b)$ and $(c, d)$ s.t. $$\operatorname{lcm}(a, b) = \operatorname{lcm}(c, d)$$ $$\gcd(a, b) = \gcd(c, d)$$ and $$\frac{a + b}{2} = \frac{c + d}{2}$$

It is easy to show that if both LCM and GCD are equal, then two pairs have the same product and the same sum AND the same GCD. I have an intuition that it is impossible that two distinct pairs can exist under these conditions but it is unclear how to strictly prove it.

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    $\begingroup$ What means lcd? LCM? $\endgroup$ – Wuestenfux Aug 20 at 11:29
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    $\begingroup$ @Wuestenfux yeah my mistake $\endgroup$ – Akhmad Sumekenov Aug 20 at 11:31
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    $\begingroup$ Or $(a,b)=(2,3)$, $(c,d)=(6,1)$ $\endgroup$ – lulu Aug 20 at 11:32
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    $\begingroup$ @lulu their means are not equal $\endgroup$ – Akhmad Sumekenov Aug 20 at 11:34
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    $\begingroup$ Well, $a+b=c+d\implies a^2+2ab+b^2=c^2+2cd+d^2$ and $ab=cd$ then implies that $(a-b)^2=(c-d)^2$. Thus $a-b=\pm(c-d)$. Now just go case by case. $\endgroup$ – lulu Aug 20 at 11:37
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Well, of course not.

This is because the product of two numbers is the product of their gcd and their lcm.

Therefore if $(a,b) = (c,d)$ and $\operatorname{lcm}[a,b] = \operatorname{lcm}[c,d]$ then taking the product gives $ab = cd$.

Furthermore you want essentially $a+b = c+d$. Squaring this , subtract the equation $4ab = 4cd$ from both sides and take the square root to get $|a-b| = |c-d|$. This results in either $a-b = c-d$ or $b-a = c-d$. Which gives that $(a,b)$ is a permutation of $(c,d)$.

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If two pairs of numbers have an LCM of $x$ and a GCD of $y$, then they have the same product (i.e. $xy$). If two pairs of numbers have a mean of $z$, then they both have the same sum (i.e. $2z$).

So, given that $\{a,b\}$ and $\{c,d\}$ have the same sum S and product P, can they be distinct? No. The graph of $x+y=S$ is a line parallel to $y=-x$, and the graph of $xy=P$ is a rectangular hyperbola. That system of equations has two solutions, and they are symmetric about the line $y=x$.

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We have $\ a\!+\!b= c\!+\!d,\ $ $\,ab = cd\,$ using $\,{\rm lcm}(x,y)\gcd(x,y) = xy$

thus $\,(x\!-\!a)(x\!-\!b) = (x\!-\!c)(x\!-\!d)\,$ have same roots so $\,\{a,b\} = \{c,d\}$

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If there exist such integers $a,b,c,d$ that satisfy the conditions, then are generated by the system: $$\left\{\begin{matrix} a+b=c+d \\ab=cd \end{matrix}\right.$$

From this, I obtai, substituing $a=\frac{cd}{b}$: $c(d-b)=b(d-b)$, so $c=b$ and fron the first equation $a=d$.

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  • $\begingroup$ $c=a$, $d=b$ is also a solution. $\endgroup$ – Ilmari Karonen Aug 20 at 21:15
  • $\begingroup$ Yes, it is: it depends in what way you solve the system. $\endgroup$ – Matteo Aug 21 at 6:31

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