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

(Created page with "==Problem 1== Find all positive integers <math>x,y</math> satisfying the equation <cmath> 9(x^2+y^2+1) + 2(3xy+2) = 2005 . </cmath> == Solution == We can re-write the equa...") |
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Line 15: | Line 15: | ||

Since <math>498 - 2y^2 \ge 0</math>. this implies that <math>y \le 15</math> | Since <math>498 - 2y^2 \ge 0</math>. this implies that <math>y \le 15</math> | ||

− | + | Also, taking <math>mod 3</math> on both sides we see that <math>y</math> cannot be a multiple of <math>3</math>. Also, note that <math>249 - y^2</math> has to be even since <math>(498 - 2y^2) = 2(249 - y^2)</math> is a perfect square. | |

+ | So, <math>y^2</math> cannot be even, implying that <math>y</math> is odd. | ||

+ | |||

+ | So we have only <math>\{1, 5, 7, 11, 13\}</math> to consider for <math>y</math>. | ||

+ | |||

+ | Trying above 5 values for <math>y</math> we find that <math>y = 7, 11</math> result in perfect squares. | ||

Thus, we have <math>2</math> cases to check: | Thus, we have <math>2</math> cases to check: | ||

Line 21: | Line 26: | ||

<math>Case 1: y = 7</math> | <math>Case 1: y = 7</math> | ||

− | <math | + | <math> (3x + 7)^2 = 4(498 - 2(7^2))</math> |

− | |||

<math> => (3x + 7)^2 = 4(400)</math> | <math> => (3x + 7)^2 = 4(400)</math> | ||

<math> => x = 11</math> | <math> => x = 11</math> | ||

Line 28: | Line 32: | ||

<math>Case 2: y = 11</math> | <math>Case 2: y = 11</math> | ||

− | <math | + | <math> (3x + 11)^2 = 4(498 - 2(11^2))</math> |

− | |||

<math> => (3x + 11)^2 = 4(256)</math> | <math> => (3x + 11)^2 = 4(256)</math> | ||

<math> => x = 7</math> | <math> => x = 7</math> |

## Revision as of 01:18, 25 December 2018

## Problem 1

Find all positive integers satisfying the equation

## Solution

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 .