# 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.

• What means lcd? LCM? Aug 20, 2019 at 11:29
• @Wuestenfux yeah my mistake Aug 20, 2019 at 11:31
• Or $(a,b)=(2,3)$, $(c,d)=(6,1)$
– lulu
Aug 20, 2019 at 11:32
• @lulu their means are not equal Aug 20, 2019 at 11:34
• 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.
– lulu
Aug 20, 2019 at 11:37

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 $$(c,d)$$.

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 $$y=x$$.

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 $$\,\{a,b\} = \{c,d\}$$

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 $$a=d$$.

• $c=a$, $d=b$ is also a solution. Aug 20, 2019 at 21:15
• Yes, it is: it depends in what way you solve the system. Aug 21, 2019 at 6:31