Perfect square and prime number A non-zero natural number $N$ is such as  $N(N+2013)$ is a perfect square.


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*Show that $N$ can not be a prime number.

*Find a $N$ value such as $N(N+2013)$ is a perfect square.
I've tried to proceed (using a proof by contradiction) assuming that $N(N+2013)$ is a perfect square and $N$ a prime number, then I decomposed $N(N+2013)$ in prime factors sum of $p_i^{2}$ and the fact that $2013$ was $3\times 11 \times 61$ but I was not able to get it.
 A: $$N(N+2013 ) =y^2 $$
$$N^2 +2\cdot \frac{2013}{2 } \cdot N +\frac{2013^2}{4} -\frac{2013^2}{4} =y^2 $$
$$4y^2 -(2N +2013)^2 =-2013^2 $$
$$(2y -2N -2013 )(2y +2N +2013 ) =-2013^2$$
and you can show the above equation using the factorization of $2013^2$
A: looking at prime factorisation of 2013 the problem becomes; $$N(N+3*11*61) = m^2$$
if we let $N$ equal the product of two of the primes and anyother number say $n$
we have $$P_1*P_2*n(P_1*P_2*(P_3 + n) = m^2$$ where $p_i$ is one of ${3,11,61}$
It is easy to see that if we take $N = 11*61*1$ we will have $$11*61*11*61*(3+1)$$
which gives $$(11*61*2)^2$$
so $m=11*61*2 = 1342$ and $N=671$
A: The answer by KingJ finds the solution when he assumes $P_3=3$. The same method can be applied to the assumptions that $P_3=11$ and $P_3=61$. There is a lot of arithmetic involved in finding the actual numbers, but they turn out to be $N=4575$ and $N=29700$.
For $N=4575$, $4575\cdot 6588=2^23^45^261^2$, a perfect square.
For $N=29700$, $29700\cdot 31713=2^23^45^211^231^2$, a perfect square.
