2013 + k is a perfect square . Find all the possible values of k.

Question

mr.math · Answer

You want to find all values of k such that
$2013+k=n^2 \implies k=n^2-2013$, which has infinitely many solutions over the integers. Thus there are infinitely many k such that $2013+k$ is a perfect square.

dumbcow · Answer

x^2 = 2013 + k

-> k = x^2 -2013, for all integers x>0

anonymous · Answer

k>=-2013

anonymous · Answer

and the smallest k is =-2012

anonymous · Answer

k=-2013 is that possible?

anonymous · Answer

(a+b+c)^2 = (2013 + k) is the original equation.

anonymous · Answer

are we looking for solution in R or C?

anonymous · Answer

In integers.

anonymous · Answer

Mr. Math and dumbcow seem right..

anonymous · Answer

a,b,c are also integers?

anonymous · Answer

yes

earthcitizen · Answer

there should be only two values of k ?

anonymous · Answer

no,you are supposed to find all the possible values .

anonymous · Answer

all possible values seem infinite..

anonymous · Answer

1 more condition : a^ + 2bc = 2012, b^2 + 2ca = 1,c^2 + 2ab = k.

anonymous · Answer

(:

mr.math · Answer

Lol, Could your write the whole problem so we can help?

anonymous · Answer

If a, b, c are integers such that  a^ + 2bc = 2012, b^2 + 2ca = 1,c^2 + 2ab = k.Then,find all the possible values of k.

anonymous · Answer

$$ac \le0 \qquad a \le0 \quad or \quad c \le0$$

anonymous · Answer

is there any options?