Perfect squares
Find the sum of all the values of x + y where x and y are positive integers such that:
x 2 + 3 x + 3 = y 2
Note: If you think that there are infinitely many solutions or no solution, type 0 .
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
exactly how I did it. You could also make it as a quadratic in x and calculate discriminant which is a perfect square
Yes. Even I did the same method Nitin Kumar!!!
Any pair ( x , ± x 2 + 3 x + 3 ) satisfy the equation. Hence there are infinite number of solutions and the answer is 0 .
wrong. It is not true that (x^2+3x+3)^1/2 is integral for infinite values of x
Log in to reply
Originally it was not mentioned that x , y are integers.
Log in to reply
oh, I see, fine
Log in to reply
@Nitin Kumar – Imran should now change the correct answer to -6
@Nitin Kumar – sorry, i replied to myself
infact only for x=-2, -1
Log in to reply
So you proved yourself that the correct answer is not 0 .
Log in to reply
yes, exactly
when i typed 0, I actually guessed
thought it would be the answer
Actually there are four solutions in integers, but no solutions in positive integers. The question has been edited to include the word positive .
We manipulate the equation as follows: x 2 + 3 x + 3 4 x 2 + 1 2 x + 1 2 ( 2 x + 3 ) 2 + 3 4 y 2 − ( 2 x + 3 ) 2 ( 2 y + 2 x + 3 ) ( 2 y − 2 x − 3 ) = y 2 = 4 y 2 = 4 y 2 = 3 = 3 so we have four possible solutions ( 2 y + 2 x + 3 , 2 y − 2 x − 3 ) = ( 3 , 1 ) , ( − 3 , − 1 ) , ( 1 , 3 ) , ( − 1 , − 3 ) which lead to four possible solutions in integers ( x , y ) = ( − 1 , 1 ) , ( − 2 , − 1 ) , ( − 2 , 1 ) , ( − 1 , − 1 ) respectively. Thus there are no solutions in positive integers, so the answer is 0 .