Simple Question On Relationship Between Cubes And Squares I'm new to this number theory business, not to mention terribly naive.  I wonder whether someone could explain the technique (assuming there is one) to show whether the expression

$12C - 3$

(where $C$ is any cube number greater than $1$) can ever be a square number.

In particular, I'm looking for a square number congruent to $3 \mod 6$, but I'll settle for just any square number for now.  A simple search of the first few hundred cube numbers turned up nothing (apart from the trivial $C = 1$)

Sorry if this question has you mathematicians rolling your eyes.

Thank you so much for your help.  By the way, if anyone can suggest a really gentle introduction to number theory, would you mind recommending a book to me?  Thanks.

Regards,

Mike.
First of all, $12C-3 = 6(2C-1)+3$, so $12C-3 \equiv 3 \pmod 6$ for any integer $C$. As Henning Makholm wrote, if a square is $3 \pmod 6$, then the root of the square is also $3 \pmod 6$, so if $12a^3-3 = k^2$, then $k = 6c+3$ for an integer $c$, and the relation can rewritten as
$$
12a^3-3 = (6c+3)^2,
$$
which is equivalent to
$$
a^3 = 3c^2+3c+1
$$
or 
$$
a^3+c^3 = (c+1)^3.
$$
This is impossible for $c&gt;0$ (it's a particular case of Fermat's Last Theorem for $n=3$; this case has been proven by Leonhard Euler), so $C=1$ is the only solution to your problem.
$1 ^ 3 + 2 ^ 3 = 3 ^ 2$. Likewise $1 ^ 3 + 2 ^ 3 + 3 ^ 3 = 6 ^ 2$. $1+2= 3$ and $1+2+3 = 6$. Like wise sum of cubes of first $4$ number is square of $10$ ($1+2+3+4$). How is it? You can check up to any number of your choice and prove whether I am right or wrong.