Difference between revisions of "2005 JBMO Problems/Problem 1"

Revision as of 13:33, 22 April 2019

Problem

Find all positive integers $x,y$ satisfying the equation $9(x^2+y^2+1) + 2(3xy+2) = 2005 .$

Solutions

Solution 1

We can re-write the equation as:

$3x^2 + y^2 + 2(3x)(y) + 8y^2 + 9 + 4 = 2005$

or $(3x + y)^2 = 4(498 - 2y^2)$

The above equation tells us that $(498 - 2y^2)$ is a perfect square. Since $498 - 2y^2 \ge 0$ . this implies that $y \le 15$

Also, taking $mod 3$ on both sides we see that $y$ cannot be a multiple of $3$ . Also, note that $249 - y^2$ has to be even since $(498 - 2y^2) = 2(249 - y^2)$ is a perfect square. So, $y^2$ cannot be even, implying that $y$ is odd.

So we have only $\{1, 5, 7, 11, 13\}$ to consider for $y$ .

Trying above 5 values for $y$ we find that $y = 7, 11$ result in perfect squares.

Thus, we have $2$ cases to check:

$Case 1: y = 7$

$(3x + 7)^2 = 4(498 - 2(7^2))$ $=> (3x + 7)^2 = 4(400)$ $=> x = 11$

$Case 2: y = 11$

$(3x + 11)^2 = 4(498 - 2(11^2))$ $=> (3x + 11)^2 = 4(256)$ $=> x = 7$

Thus all solutions are $(7, 11)$ and $(11, 7)$ .

$Kris17$

Solution 2

Expanding, combining terms, and factoring results in $\begin{align*} 9x^2 + 9y^2 + 9 + 6xy + 4 &= 2005 \\ 9x^2 + 9y^2 + 6xy &= 1992 \\ 3x^2 + 3y^2 + 2xy &= 664 \\ (x+y)^2 + 2x^2 + 2y^2 &= 664. \end{align*}$ Since $2x^2$ and $2y^2$ are even, $(x+y)^2$ must also be even, so $x$ and $y$ must have the same parity. There are two possible cases.

Case 1: $x$ and $y$ are both even

Let $x = 2a$ and $y = 2b$ . Substitution results in $\begin{align*} 4(a+b)^2 + 8a^2 + 8b^2 &= 664 \\ (a+b)^2 + 2a^2 + 2b^2 &= 166 \end{align*}$ Like before, $a+b$ must be even for the equation to be satisfied. However, if $a+b$ is even, then $(a+b)^2$ is a multiple of 4. If $a$ and $b$ are both even, then $2a^2 + 2b^2$ is a multiple of 4, but if $a$ and $b$ are both odd, the $2a^2 + 2b^2$ is also a multiple of 4. However, $166$ is not a multiple of 4, so there are no solutions in this case.

Case 2: $x$ and $y$ are both odd

Let $x = 2a+1$ and $y = 2b+1$ , where $a,b \ge 0$ . Substitution and rearrangement results in $\begin{align*} 4(a+b+1)^2 + 2(2a+1)^2 + 2(2b+1)^2 &= 664 \\ 2(a+b+1)^2 + (2a+1)^2 + (2b+1)^2 &= 332 \\ 6a^2 + 4ab + 6b^2 + 8a + 8b &= 328 \\ 3a^2 + 2ab + 3b^2 + 4a + 4b &= 164 \end{align*}$ Note that $3a^2 \le 164$ , so $a \le 7$ . There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in $3b^2 + b(2a+4) + 3a^2 + 4a - 164 = 0$ . Since both $a$ and $b$ are integers, we must have $\begin{align*} n^2 &= 4a^2 + 16a + 16 - 12(3a^2 + 4a - 164) \\ &= -32a^2 - 32a + 16 + 12 \cdot 164 \\ &= 16(-2a^2 - 2a + 1 + 3 \cdot 41) \\ &= 16(-2(a^2 + a) + 124), \end{align*}$ where $n$ is an integer. Thus, $-2(a^2 + a) + 124$ must be a perfect square.

After trying all values of $a$ from 0 to 7, we find that $a$ can be $3$ or $5$ . If $a = 3$ , then $b = 5$ , and if $a = 5$ , then $b = 3$ .

Therefore, the ordered pairs $(x,y)$ that satisfy the original equation are $\boxed{(7,11) , (11,7)}$ .

@@ Line 1: / Line 1: @@
-==Problem 1==
+==Problem==
 Find all positive integers <math>x,y</math> satisfying the equation <cmath> 9(x^2+y^2+1) + 2(3xy+2) = 2005 . </cmath>
+== Solutions ==
-== Solution ==
+===Solution 1===
 We can re-write the equation as:
@@ Line 40: / Line 41: @@
 <math>Kris17</math>
+===Solution 2===
+Expanding, combining terms, and factoring results in
+<cmath>\begin{align*}
+x^2 + 9y^2 + 9 + 6xy + 4 &= 2005 \\
+x^2 + 9y^2 + 6xy &= 1992 \\
+x^2 + 3y^2 + 2xy &= 664 \\
+(x+y)^2 + 2x^2 + 2y^2 &= 664.
+\end{align*}</cmath>
+Since <math>2x^2</math> and <math>2y^2</math> are even, <math>(x+y)^2</math> must also be even, so <math>x</math> and <math>y</math> must have the same parity.  There are two possible cases.
+'''Case 1: <math>x</math> and <math>y</math> are both even'''
+Let <math>x = 2a</math> and <math>y = 2b</math>.  Substitution results in
+<cmath>\begin{align*}
+(a+b)^2 + 8a^2 + 8b^2 &= 664 \\
+(a+b)^2 + 2a^2 + 2b^2 &= 166
+\end{align*}</cmath>
+Like before, <math>a+b</math> must be even for the equation to be satisfied.  However, if <math>a+b</math> is even, then <math>(a+b)^2</math> is a multiple of 4.  If <math>a</math> and <math>b</math> are both even, then <math>2a^2 + 2b^2</math> is a multiple of 4, but if <math>a</math> and <math>b</math> are both odd, the <math>2a^2 + 2b^2</math> is also a multiple of 4.  However, <math>166</math> is not a multiple of 4, so there are no solutions in this case.
+'''Case 2: <math>x</math> and <math>y</math> are both odd'''
+Let <math>x = 2a+1</math> and <math>y = 2b+1</math>, where <math>a,b \ge 0</math>.  Substitution and rearrangement results in
+<cmath>\begin{align*}
+(a+b+1)^2 + 2(2a+1)^2 + 2(2b+1)^2 &= 664 \\
+(a+b+1)^2 + (2a+1)^2 + (2b+1)^2 &= 332 \\
+a^2 + 4ab + 6b^2 + 8a + 8b &= 328 \\
+a^2 + 2ab + 3b^2 + 4a + 4b &= 164
+\end{align*}</cmath>
+Note that <math>3a^2 \le 164</math>, so <math>a \le 7</math>.  There are only a few cases to try out, so we can do guess and check.  Rearranging terms once more results in <math>3b^2 + b(2a+4) + 3a^2 + 4a - 164 = 0</math>.  Since both <math>a</math> and <math>b</math> are integers, we must have
+<cmath>\begin{align*}
+n^2 &= 4a^2 + 16a + 16 - 12(3a^2 + 4a - 164) \\
+&= -32a^2 - 32a + 16 + 12 \cdot 164 \\
+&= 16(-2a^2 - 2a + 1 + 3 \cdot 41) \\
+&= 16(-2(a^2 + a) + 124),
+\end{align*}</cmath>
+where <math>n</math> is an integer.  Thus, <math>-2(a^2 + a) + 124</math> must be a perfect square.
+<br>
+After trying all values of <math>a</math> from 0 to 7, we find that <math>a</math> can be <math>3</math> or <math>5</math>.  If <math>a = 3</math>, then <math>b = 5</math>, and if <math>a = 5</math>, then <math>b = 3</math>.
+<br>
+Therefore, the ordered pairs <math>(x,y)</math> that satisfy the original equation are <math>\boxed{(7,11) , (11,7)}</math>.
+==See Also==
+{{JBMO box|year=2005|before=First Problem|num-a=2|five=}}
+[[Category:Intermediate Number Theory Problems]]

2005 JBMO (Problems • Resources)
Preceded by First Problem	Followed by Problem 2
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