2005 JBMO Problems/Problem 1

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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 $\boxed{(7,11) , (11,7)}$.

See Also

2005 JBMO (ProblemsResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4
All JBMO Problems and Solutions
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