Difference between revisions of "1980 AHSME Problems/Problem 29"

Latest revision as of 01:03, 26 June 2025

Problem

How many ordered triples $(x,y,z)$ of integers satisfy the system of equations below?

$\begin{align*} x^2 - 3xy + 2y^2 - z^2 &= 31 \\ -x^2 + 6yz + 2z^2 &= 44 \\ x^2 + xy + 8z^2 &= 100 \end{align*}$

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ \text{a finite number greater than 2}\qquad \text{(E)}\ \text{infinitely many}$

Solution 1 (Sum of Squares)

We are considering the sum of three equations:

$x^2 - 2xy + 2y^2 + 6yz + 9z^2 = (x - y)^2 + (y + 3z)^2 = 175.$

Since $x, y, z$ are integers, this implies that $175$ can be written as the sum of two squares, i.e.,

$175 = a^2 + b^2,$

where $a$ and $b$ are integers.

However, $175 \equiv 3\ (\mathrm{mod}\ 4)$ , but $a^2$ and $b^2$ must each be congruent to either $0$ or $1 (\mathrm{mod}\ 4)$ . So this equation, and therefore this system, has $0$ integer solutions. $\fbox{(A)}$

Wwei.yu (talk) 22:09, 28 March 2020 (EDT)

Solution 2 (Eliminate Cross Terms)

First, to eliminate the $xy$ term, multiply both sides of the third equation by $3$ to get $3x^2 + 3xy + 24z^2 = 300$ , and then add it to the first equation to get $4x^2 + 2yz + 23z^2 = 331$ .

Next, to eliminate the $yz$ term, multiply both sides of the new equation by $-3$ to get $-12x^2 - 6yz - 69z^2 = -993$ , and then add it to the second equation to get $-13x^2 - 67z^2 = -949$ .

This equation can be rearranged to get $67z^2 = 949-13x^2 = 13(73-x^2)$ . Since $67$ is not divisible by $13$ , $z$ must be divisible by $13$ and there exists an integer $n$ such that $z = 13n$ . Substituting $z = 13n$ and dividing both sides of this equation by $13$ yields $871n^2 = 73 - x^2$ . So $871n^2 \leq 73$ , which is only true if $n = 0$ . But $n = 0$ would require $x^2 = 73$ , which is impossible since $73$ is not a perfect square. Therefore, this equation (and also this system) has $0$ integer solutions. $\fbox{(A)}$

-j314andrews, based on solution by Farenhajt

Solution 3 (Modular Arithmetic)

Reducing the second equation mod $2$ yields $x^2 \equiv 0\ (\mathrm{mod}\ 2)$ . Therefore, $x \equiv 0\ (\mathrm{mod}\ 2)$ and $x$ is even.

Reducing the first equation mod $2$ and substituting yields $z^2 \equiv 1\ (\mathrm{mod}\ 2)$ . Therefore, $z \equiv 1\ (\mathrm{mod}\ 2)$ and $z$ is odd.

Since $x$ is even, $x^2 \equiv 0\ (\mathrm{mod}\ 4)$ , and since $z$ is odd, $z^2 \equiv 1\ (\mathrm{mod}\ 4)$ . Also $6z \equiv 2z \equiv 2\ (\mathrm{mod}\ 4)$ . Reducing the second equation mod $4$ and substituting yields $2y + 2 \equiv 0\ (\mathrm{mod}\ 4)$ , so $y$ is odd.

Reducing the third equation mod $4$ and substituting yields $xy \equiv 0\ (\mathrm{mod}\ 4)$ . Since $y$ is odd, $x \equiv 0\ (\mathrm{mod}\ 4)$ .

Since $y$ is odd, $y^2 \equiv 1\ (\mathrm{mod}\ 4)$ . Reducing the first equation mod $4$ and substituting yields $1 \equiv 3\ (\mathrm{mod}\ 4)$ , which is impossible.

Therefore, this system has $0$ integer solutions. $\fbox{(A)}$

-j314andrews

@@ Line 1: / Line 1: @@
 == Problem ==
-How many ordered triples (x,y,z) of integers satisfy the system of equations below?
+How many ordered triples <math>(x,y,z)</math> of integers satisfy the system of equations below?
-<cmath>\begin{array}{l} x^2-3xy+2y^2-z^2=31 \\ -x^2+6yz+2z^2=44 \\ x^2+xy+8z^2=100\\ \end{array} </cmath>
+<cmath>
+\begin{align*}
+x^2 - 3xy + 2y^2 - z^2 &= 31 \\
+-x^2 + 6yz + 2z^2 &= 44 \\
+x^2 + xy + 8z^2 &= 100
+\end{align*}
+</cmath>
-<math>\text{(A)} \ 0 \qquad
+<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ \text{a finite number greater than 2}\qquad \text{(E)}\ \text{infinitely many}</math>
-\text{(B)} \ 1 \qquad
-\text{(C)} \ 2 \qquad \\
-\text{(D)}\ \text{a finite number greater than 2}\qquad\\
-\text{(E)}\ \text{infinitely many} </math>
-== Solution ==
+== Solution 1 (Sum of Squares) ==
-<math>\fbox{}</math>
+We are considering the sum of three equations:
-== See also ==
+<cmath>
+x^2 - 2xy + 2y^2 + 6yz + 9z^2 = (x - y)^2 + (y + 3z)^2 = 175.
+</cmath>
+Since <math>x, y, z</math> are integers, this implies that <math>175</math> can be written as the sum of two squares, i.e.,
+<cmath>
+= a^2 + b^2,
+</cmath>
+where <math>a</math> and <math>b</math> are integers.
+However, <math>175 \equiv 3\ (\mathrm{mod}\ 4)</math>, but <math>a^2</math> and <math>b^2</math> must each be congruent to either <math>0</math> or <math>1 (\mathrm{mod}\ 4)</math>. So this equation, and therefore this system, has <math>0</math> integer solutions. <math>\fbox{(A)}</math>
+[[User:Wwei.yu|Wwei.yu]] ([[User talk:Wwei.yu|talk]]) 22:09, 28 March 2020 (EDT)
+== Solution 2 (Eliminate Cross Terms) ==
+First, to eliminate the <math>xy</math> term, multiply both sides of the third equation by <math>3</math> to get <math>3x^2 + 3xy + 24z^2 = 300</math>, and then add it to the first equation to get <math>4x^2 + 2yz + 23z^2 = 331</math>.
+Next, to eliminate the <math>yz</math> term, multiply both sides of the new equation by <math>-3</math> to get <math>-12x^2 - 6yz - 69z^2 = -993</math>, and then add it to the second equation to get <math>-13x^2 - 67z^2 = -949</math>.
+This equation can be rearranged to get <math>67z^2 = 949-13x^2 = 13(73-x^2)</math>.  Since <math>67</math> is not divisible by <math>13</math>, <math>z</math> must be divisible by <math>13</math> and there exists an integer <math>n</math> such that <math>z = 13n</math>.  Substituting <math>z = 13n</math> and dividing both sides of this equation by <math>13</math> yields <math>871n^2 = 73 - x^2</math>.  So <math>871n^2 \leq 73</math>, which is only true if <math>n = 0</math>.  But <math>n = 0</math> would require <math>x^2 = 73</math>, which is impossible since <math>73</math> is not a perfect square.  Therefore, this equation (and also this system) has <math>0</math> integer solutions. <math>\fbox{(A)}</math>
+-j314andrews, based on solution by Farenhajt
+== Solution 3 (Modular Arithmetic) ==
+Reducing the second equation mod <math>2</math> yields <math>x^2 \equiv 0\ (\mathrm{mod}\ 2)</math>.  Therefore, <math>x \equiv 0\ (\mathrm{mod}\ 2)</math> and <math>x</math> is even.
+Reducing the first equation mod <math>2</math> and substituting yields <math>z^2 \equiv 1\ (\mathrm{mod}\ 2)</math>.  Therefore, <math>z \equiv 1\ (\mathrm{mod}\ 2)</math> and <math>z</math> is odd.
+Since <math>x</math> is even, <math>x^2 \equiv 0\ (\mathrm{mod}\ 4)</math>, and since <math>z</math> is odd, <math>z^2 \equiv 1\ (\mathrm{mod}\ 4)</math>. Also <math>6z \equiv 2z \equiv 2\ (\mathrm{mod}\ 4)</math>. Reducing the second equation mod <math>4</math> and substituting yields <math>2y + 2 \equiv 0\ (\mathrm{mod}\ 4)</math>, so <math>y</math> is odd.
+Reducing the third equation mod <math>4</math> and substituting yields <math>xy \equiv 0\ (\mathrm{mod}\ 4)</math>.  Since <math>y</math> is odd, <math>x \equiv 0\ (\mathrm{mod}\ 4)</math>.
+Since <math>y</math> is odd, <math>y^2 \equiv 1\ (\mathrm{mod}\ 4)</math>.  Reducing the first equation mod <math>4</math> and substituting yields <math>1 \equiv 3\ (\mathrm{mod}\ 4)</math>, which is impossible.
+Therefore, this system has <math>0</math> integer solutions. <math>\fbox{(A)}</math>
+-j314andrews
+== See Also ==
 {{AHSME box|year=1980|num-b=28|num-a=30}}
+{{MAA Notice}}
 [[Category: Intermediate Algebra Problems]]
-{{MAA Notice}}

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

Page

Toolbox

Search