<s>[INST] Perfect Squares ending in 576 I want to find out perfect squares ending in 576, after the number 576.
Here is my derivation to arrive at such a number. Let the perfect square ending in $576$ be $1000k+576$. Every perfect square can be expressed as a the sum of a certain number of consecutive odd numbers. For eg: $2^2 = 1+3$, $3^2 = 1+3+5$, $4^2 = 1+3+5+7$, and so on..
Hence I can write my required perfect square ending in 576 as -
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{(n\text{ summands})}$$
Therefore, 
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{
(n \text{ summands})} = 1000k +576.$$ 
Since $(1+3+5+7+ ....49) = 576$, the equation reduces to
$$\underbrace{(51+53+55+57+\cdots)}_{n\text{ summands})} = 1000k$$
Using formula for Arithmetic Progression starting with 51 and a common difference of 2,
$$\begin{align*}
\frac{n}{2}\left(2(51) + (n-1)2\right) &= 1000k\\
n(n+50) &= 1000k
\end{align*}$$
Put $n = 100$, $100\times 150 = 1000k$, hence $k = 15$.
Put $k = 15$ in the perfect square term $1000k+576$ we get the number $15576$.
But $15576$ is not a perfect square.
What is flawed in my derivation? Kindly help.
 [/INST] Your question definitely deserves a more precise answer than this, see the answer by Peter Taylor.
But I'd like to point out that what you are looking for are squares that are $576 \pmod{1000}$, i.e. solutions to $x^2 = 24^2 \pmod{1000}$.
This reduces to the equations
$$ \begin{align*} x^2 &= 24^2 \pmod{125} \\ x^2 &= 24^2 \pmod{8} \end{align*}. $$
The first solution can be solved to $x \equiv \pm 24 \pmod{125}$; the second implies $x\equiv 0\pmod 4$. We may then use the Chinese Remainder Theorem to reassemble these and we find that $x \mod 1000 \in \{24,476,524,976\}$ (This comes from: $x \mod 500 = \pm 24$.) Thus the squares ending in $576$ are exacly the squares of the numbers ending in 024, 476, 524 and 976. For instance $1024^2 = 1048576$.
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