Prove two distinct pairs of natural numbers with these properties do not exist 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.
 A: 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 $(c,d)$.
A: 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 $y=x$.
A: 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 $\,\{a,b\} = \{c,d\}$ 
A: 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 $a=d$.
