Pair me up
Find the number of ordered pairs of integers ( x , y ) such that x 2 − 3 x y + 2 y 2 = 2 7 .
Details and assumptions
For an ordered pair of integers ( a , b ) , the order of the integers matter. The ordered pair ( 1 , 2 ) is different from the ordered pair ( 2 , 1 ) .
The answer is 8.
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5 solutions
Moderator note:
You have merely found a necessary condition for a solution, i.e. if a solution exists it must satisfy certain properties.
You have not established that it is a sufficient solution, i.e. that the pairs you claim are indeed (integer) solutions.
x − y = m ... ( 1 )
x − 2 y = n ... ( 2 )
Solving for x in ( 1 ) and substituting in ( 2 ) gives:
y = m − n
Substituting y in ( 2 ) :
x = 2 m − n
This two equations show that x and y are integers because of m and n are integers too.
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Great!
If the simultaneous equations were more complicated, then just because m and n are integers does not guarantee that x and y are integers. For example, in the system
{ 2 x + y = m x + 2 y = n
Then x and y are integers if and only if m + n is a multiple of 3.
We can factor the left hand side:
x 2 − 2 x y + y 2 + y 2 − x y = ( x − y ) 2 − y ( x − y ) = ( x − y ) ( x − 2 y )
Clearly, x − y and x − 2 y are factors of 2 7 . We can divide 2 7 into pairs of factors as follows: 1 × 2 7 , 3 × 9 , ( − 1 ) × ( − 2 7 ) , ( − 3 ) × ( − 9 ) .
Each pair of factors will yield two solutions since we can set x equal to either of the factors and then solve for y . All of these will be distinct. Therefore, the answer is 8 pairs.
Note that x 2 − 3 x y + 2 y 2 = 2 7 can be expressed by ( x − y ) ( x − 2 y ) = 2 7 . We should solve 8 possibilities. For each one:
x − y = @ = > 2 x − 2 y = 2 @
x − 2 y = $
Here, @ and $ are numbers for each possibility.
Subtracting, we have x = 2 @ − $ , giving 8 possibilities for ( x ).
Moderator note:
This comes close to explaining why the 8 pairs of factors of 27 leads to 8 integer solutions. How would you present this argument better?
The polynomial can be factorized as ( x − 2 y ) ( x − y ) = 2 7
27 can be represented as a product of two integers in 8 ways ( 9 × 3 , 3 × 9 , 2 7 × 1 , 1 × 2 7 , − 3 × − 9 , − 9 × − 3 , − 2 7 × − 1 , − 1 × − 2 7 ) and each will give us a different ordered pair.So the answer is 8.
By factoring the expression x 2 -3xy + 2 y 2 = (x-2y)(x-y) Lets focus on 27 for the factors including negative pairs. {1,27},{27,1},{3,9},{9,3},{-1,-27},{-27,-1},{-3,-9},{-9,-3}
By doing pair in pair ,. for example [x-2y=1,x-y=27] we can solve for the value of x and y. Do it in the other factors. We notice that for all values of x and y, there are all integer ordered pairs. Therefore there are 8.
there was any assumptions that there is negative ordered pairs ????
x 2 − 3 x y + 2 y 2 = ( x − y ) ( x − 2 y ) = 2 7
( x − y ) = m
( x − 2 y ) = n
m n = 2 7 = 3 3 and so 27 has 4 positive divisors ( 1 , 3 , 9 , 2 7 ) .
We have to count them twice because m and n can be both negative.
So we have 8 pairs.