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

Revision as of 22:09, 25 March 2023

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

Let $a=dzxA$ $b=dxyB$ $c=dyzC$ where $\gcd(A,B,C)=\gcd(x,y,z)=1$ . (Existence proof missing.) 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.

@@ Line 47: / Line 47: @@
 Finally, canceling out the <math>y_i</math>'s and collecting like terms, we see the RHS is <math>2z_i - 2x_i</math>, which is equal to the LHS. Thus, this equation is true, and we are done. <math>\square</math>
-===Solution 4, Very Simple===
+===Solution 4===
 Let
-<cmath>a=dzxA</cmath><cmath>b=dxyB</cmath><cmath>c=dyzC</cmath>where <math>\gcd(A,B,C)=\gcd(x,y,z)=1</math>. Then, <math>\operatorname{lcm}(a,b,c)^2=(dxyzABC)^2</math> and <math>\operatorname{lcm}(a,b)=dxyzAB</math>. Using this cyclically, we have
+<cmath>a=dzxA</cmath><cmath>b=dxyB</cmath><cmath>c=dyzC</cmath>where <math>\gcd(A,B,C)=\gcd(x,y,z)=1</math>. (Existence proof missing.) Then, <math>\operatorname{lcm}(a,b,c)^2=(dxyzABC)^2</math> and <math>\operatorname{lcm}(a,b)=dxyzAB</math>. Using this cyclically, we have
 <cmath>\text{LHS}=\frac{d^2x^2y^2z^2A^2B^2C^2}{(dxyzAB)(dxyzBC)(dxyzCA)}=\frac{1}{dxyz}.</cmath>Now, note that <math>\gcd(a,b,c)^2=d^2</math> and <math>\gcd(a,b)=dx</math>. Using this cyclically, we have
 <cmath>\text{RHS}=\frac{d^2}{(dx)(dy)(dz)}=\frac{1}{dxyz}.</cmath>

1972 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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