# Difference between revisions of "1972 USAMO Problems/Problem 1"

## Problem

The symbols $(a,b,\ldots,g)$ and $[a,b,\ldots, g]$ denote the greatest common divisor and least common multiple, respectively, of the positive integers $a,b,\ldots, g$. For example, $(3,6,18)=3$ and $[6,15]=30$. Prove that

$\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$.

## Solutions

### Solution 1

As all of the values in the given equation are positive, we can invert it to get an equivalent equation:

$\frac{[a,b][b,c][c,a]}{[a,b,c]^2}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}$.

We will now prove that both sides are equal, and that they are integers.

Consider an arbitrary prime $p$. Let $p^\alpha$, $p^\beta$, and $p^\gamma$ be the greatest powers of $p$ that divide $a$, $b$, and $c$. WLOG let $\alpha \leq \beta \leq \gamma$.

We can now, for each of the expressions in our equation, easily determine the largest power of $p$ that divides it. In this way we will find that the largest power of $p$ that divides the left hand side is $\beta+\gamma+\gamma-2\gamma = \beta$, and the largest power of $p$ that divides the right hand side is $\alpha + \beta + \alpha - 2\alpha = \beta$. $\blacksquare$

### Solution 2

Let $p = (a, b, c)$, $pq = (a, b)$, $pr = (b, c)$, and $ps = (c, a)$. Then it follows that $q, r, s$ are pairwise coprime and $a = pqsa'$, $b = pqrb'$, and $c = prsc'$, with $a', b', c'$ pairwise coprime as well. Then, we wish to show $$\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)},$$ which can be checked fairly easily.

### Solution 3 (very long, but detailed)

Note: This solution is more of a "bashy" solution, and is easier to think of than the first 2 solutions.

Dividing both sides of the equation by $(a,b,c)^2$, and then multiplying both sides by $[a,b][b,c][c,a]$ gives us $$\left(\frac{[a,b,c]}{(a,b,c)}\right)^2 = \frac{[a,b][b,c][c,a]}{(a,b)(b,c)(c,a)}.$$ Now, we look at the prime factorisations of $a,b,c$. Let $$a=2^{x_1}\cdot 3^{x_2}\cdot 5^{x_3}\cdots,$$ $$b=2^{y_1}\cdot 3^{y_2}\cdot 5^{y_3}\cdots,$$ $$c=2^{z_1}\cdot 3^{z_2}\cdot 5^{z_3}\cdots,$$ where all of the $x_1, y_1, z_1$ are nonnegative integers (they could be 0). Thus, we have $$[a,b,c]=2^{\max(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3)} \cdots$$ and $$(a,b,c)=2^{\min(x_1,y_1,z_1)} \cdot 3^{\min(x_2,y_2,z_2)} \cdot 5^{\min(x_3,y_3,z_3)} \cdots.$$ Now, we see that $\left(\frac{[a,b,c]}{(a,b,c)}\right) = 2^{\max(x_1,y_1,z_1) - \min(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2) - \min(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3) - \min(x_3,y_3,z_3)} \cdots.$ Squaring the LHS will just double all the exponents on the RHS.

Also, we have $[a,b]=2^{\max(x_1,y_1)} \cdot 3^{\max(x_2,y_2)} \cdot 5^{\max(x_3,y_3)} \cdots$. We can also build similar equations for $[b,c],[c,a],(a,b),(b,c)(c,a)$, using 2 of $x,y,z$ and either the minimum of the exponents or the maximum. We see that when we multiply $[a,b]$, $[b,c]$, and $[a,c]$ together, the exponents of the prime factorization will be $\max(x_i,y_i)+ \max(y_i,z_i)+ \max(z_i,x_1)$, where $i$ is chosen for the $i$'th prime. When we multiply $(a,b)$, $(b,c)$, and $(a,c)$ together, the exponents of the prime factorization will be $\min(x_i,y_i)+ \min(y_i,z_i)+ \min(z_i,x_1)$, where $i$ is chosen for the $i$'th prime.

Thus, when we divide the former by the latter, we have $\max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))$. We wish to prove that this is equal to the exponents we got from $([a,b,c]/(a,b,c))^2$.

In other words, this problem is now down to proving that $$2(\max(x_i,y_i,z_i) - \min(x_i,y_i,z_i)) = \max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))$$ for any nonnegative integers $x_i, y_i, z_i$. WLOG, let $x_i \le y_i \le z_i$. Thus, computing maximums and minimums, this equation turns into $$2(z_i - x_i) = y_i + z_i + z_i - x_i - y_i - x_i.$$

Finally, canceling out the $y_i$'s and collecting like terms, we see the RHS is $2z_i - 2x_i$, which is equal to the LHS. Thus, this equation is true, and we are done. $\square$

### Solution 4, Very Simple

Let $$a=dzxA$$$$b=dxyB$$$$c=dyzC$$where $\gcd(A,B,C)=\gcd(x,y,z)=1$. Then, $\operatorname{lcm}(a,b,c)^2=(dxyzABC)^2$ and $\operatorname{lcm}(a,b)=dxyzAB$. Using this cyclically, we have $$\text{LHS}=\frac{d^2x^2y^2z^2A^2B^2C^2}{(dxyzAB)(dxyzBC)(dxyzCA)}=\frac{1}{dxyz}.$$Now, note that $\gcd(a,b,c)^2=d^2$ and $\gcd(a,b)=dx$. Using this cyclically, we have $$\text{RHS}=\frac{d^2}{(dx)(dy)(dz)}=\frac{1}{dxyz}.$$ -brainiacmaniac31 Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.