Find all integer solutions
How many ordered pairs of integers solutions ( x , y ) are there to 2 x 2 + 3 x y + y 2 + 2 x + y + 1 8 = 0 ?
The answer is 12.
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6 solutions
Moderator note:
Nicely done! Please read the author's comment that completes the solution, without having to consider any cases.
Note: x and y are always integers because,for each pair of divisors we have the simultaneous equation:(where m and n are the divisors of − 1 8 ):
2 x + y = m ...(1)
x + y + 1 = n ...(2)
we can solve it through substitution:solving for y in the first equation and substituting y in the second equation gives us :
x = m − n + 1
y = − m + 2 n − 2
Both the equations show that x and y must be integers as long as m and n are integers.
2 x 2 + 3 x y + y 2 + 2 x + y + 1 8 = ( 2 x + y ) ( x + y + 1 ) + 1 8
So ( 2 x + y ) ( x + y + 1 ) = − 1 8
Since { x , y } ⊂ Z we can say that ( 2 x + y , x + y + 1 ) ∈ { ( − 1 8 , 1 ) , ( − 9 , 2 ) , ( − 6 , 3 ) , ( − 3 , 6 ) , ( − 2 , 9 ) , ( − 1 , 1 8 ) , ( 1 , − 1 8 ) , ( 2 , 9 ) , ( 3 , − 6 ) , ( 6 , − 3 ) , ( 9 , − 2 ) , ( 1 8 , − 1 ) }
Through simultaneous equations and checking every case, we get the solutions of x a n d y
( − 1 8 , 1 8 ) ( − 1 0 , 1 1 ) ( − 8 , 1 0 ) ( − 8 , 1 3 ) ( − 1 0 , 1 8 ) ( − 1 8 , 3 5 ) ( 2 0 , − 3 9 ) ( 1 2 , − 2 2 ) ( 1 0 , − 1 7 ) ( 1 0 , − 1 4 ) ( 1 2 , − 1 5 ) ( 2 0 , − 2 2 )
So there are 1 2 ordered integer solutions of ( x , y )
Moderator note:
Nicely done!
the original formula
2 x 2 + 3 x y + y 2 + 2 x + y + 1 8 = 0
rearranged into
( 2 x + y ) × ( x + y ) + ( 2 x + y ) = − 1 8
( 2 x + y ) × ( x + y + 1 ) = − 1 8
let
2 x + y = m
x + y + 1 = n
so
x = m − n + 1
y = − m + 2 n − 2
we can conclude that
for all integers ( m , n )
( x , y ) will always remain integers
we known that
1 8 = 2 1 × 3 2
so 1 8 have ( 1 + 1 ) × ( 2 + 1 ) = 6 positive factors
so ( m , n ) can take on 1 2 different pairs of integers
same goes to ( x , y )
so we have 1 2 different ordered pairs of integers solutions ( x , y )
Moderator note:
This solution is basically the same as Thaddeus A.'s one. Side note: instead of using the # symbol that makes the formulas real big, just enclose them in the ∖ [ . . . ∖ ] brackets, instead of the ∖ ( . . . ∖ ) brackets.
Let's rewrite a bit...
( 2 x 2 + 3 x y + y 2 ) + ( 2 x + y ) + 1 8 = 0
⟹ ( x + y ) ( 2 x + y ) + ( 2 x + y ) = − 1 8
⟹ ( 2 x + y ) ( x + y + 1 ) = − 1 8
So, the product of ( 2 x + y ) and ( x + y + 1 ) is − 1 8 ... Let's factor out − 1 8 ...
− 1 8 = 1 × − 1 8 = − 1 × 1 8 = 2 × − 9 = − 2 × 9 = 3 × − 6 = − 3 × 6 = 6 × − 3 = − 6 × 3 = 9 × − 2 = − 9 × 2 = 1 8 × − 1 = − 1 8 × 1
These are also the possible integer values of ( 2 x + y ) and ( x + y + 1 ) ... We can see that all these values yield integer solutions of x and y ...
Since there are 1 2 ordered pairs of the factors of − 1 8 , and all these yield integer solutions of x and y , hence there are also 1 2 ordered pairs of integer solutions ( x , y ) exists...
Therefore, the required answer is 1 2
A well written solution! A small suggestion though. You could have saved yourself the trouble by not going through all the cases where − 1 8 could be expressed through the product of two integers.
You could have factored 1 8 as 2 1 ⋅ 3 2 and come to conclusion that 1 8 had ( 1 + 1 ) × ( 2 + 1 ) = 6 positive divisors and 6 negative divisors.
It would have made solution a bit cleaner and easier on the eye.
You can read about the Divisor Function here .
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Ya, you're right... That would make the solution cleaner... Thanks for the suggestion... I will keep that in mind in future while finding out the divisors of a number... Also thanks for the link... :)
2x^2 + 3xy + y^2 + 2x + y +18 = 0
x^2 + 2xy + y^2 + x^2 + 2x + 1 + xy + y +17 = 0
(x+y)^2 + (x+1)^2 + y(x+1) + 17 = 0
(x+y)^2 + (x+y+1) (x+1) + 17 = 0
LET (x+y) = k then,
k^2 + (k+1) ( k+1-y) +17 = 0
k^2 + 17 = (k+1) (y-k-1)
( k^2 + 17 ) / (k+1) = (y-k-1)
(k^2 + 2k + 1 + 16 - 2k) / (k+1) = (y-k-1)
(k+1) + (18-2k-2)/(k+1) = (y-k-1)
(k+1) + 18/(k+1) - 2 = (y-k-1)
Since x and y are integers , (y-k-1) must also be integer and it is integer only if 18/(k+1) is integer...
Number of positive factors of 18 is 6. So, number of solutions of k is 6*2=12 Thus, there are 12 ordered pairs of integer solutions
Moderator note:
One also has to show that these simultaneous solution give integer values for x and (y.)
we can write:
3xy + 2(x + (1/2))² + (y + (1/2))² + 69/4 = 0
integer solution:
(-18,18) , (-18, 35), (-10,11) , (-10,18), (-8,10)
we get 12 pairs of integer solutions
2 x 2 + 3 x y + y 2 + 2 x + y + 1 8 = 0
Factorization the first few terms gives
( 2 x + y ) ( x + y ) + 2 x + y + 1 8 = 0
( 2 x + y ) ( x + y ) + 2 x + y = − 1 8
( 2 x + y ) ( x + y + 1 ) = − 1 8
Let m = 2 x + y and n = x + y + 1 .
m n = − 1 8
From the above,it is obvious that the number of solutions are the number of pairs that when multiplied give -18.This is essentialy finding the number of divisors of − 1 8 .
Using the divisor function : from 3 2 . 2 we get ( 2 + 1 ) ( 1 + 1 ) = 6 divisors of 1 8 . We multiply this result by two because we must also include negative divisors,thus 6 . 2 = 1 2 solutions.