Difference between revisions of "2005 JBMO Problems/Problem 1"
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− | ==Problem | + | ==Problem== |
Find all positive integers <math>x,y</math> satisfying the equation <cmath> 9(x^2+y^2+1) + 2(3xy+2) = 2005 . </cmath> | 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: | We can re-write the equation as: | ||
− | <math> 3x^2 + y^2 + 2(3x)(y) + 8y^2 + 9 + 4 = 2005</math> | + | <math> (3x)^2 + y^2 + 2(3x)(y) + 8y^2 + 9 + 4 = 2005</math> |
or <math> (3x + y)^2 = 4(498 - 2y^2)</math> | or <math> (3x + y)^2 = 4(498 - 2y^2)</math> | ||
Line 40: | Line 41: | ||
<math>Kris17</math> | <math>Kris17</math> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | Expanding, combining terms, and factoring results in | ||
+ | <cmath>\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*}</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*} | ||
+ | 4(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*} | ||
+ | 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*}</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]] |
Latest revision as of 13:34, 22 April 2019
Problem
Find all positive integers satisfying the equation
Solutions
Solution 1
We can re-write the equation as:
or
The above equation tells us that is a perfect square. Since . this implies that
Also, taking on both sides we see that cannot be a multiple of . Also, note that has to be even since is a perfect square. So, cannot be even, implying that is odd.
So we have only to consider for .
Trying above 5 values for we find that result in perfect squares.
Thus, we have cases to check:
Thus all solutions are and .
Solution 2
Expanding, combining terms, and factoring results in Since and are even, must also be even, so and must have the same parity. There are two possible cases.
Case 1: and are both even
Let and . Substitution results in Like before, must be even for the equation to be satisfied. However, if is even, then is a multiple of 4. If and are both even, then is a multiple of 4, but if and are both odd, the is also a multiple of 4. However, is not a multiple of 4, so there are no solutions in this case.
Case 2: and are both odd
Let and , where . Substitution and rearrangement results in Note that , so . There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in . Since both and are integers, we must have where is an integer. Thus, must be a perfect square.
After trying all values of from 0 to 7, we find that can be or . If , then , and if , then .
Therefore, the ordered pairs that satisfy the original equation are .
See Also
2005 JBMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |