# Perfect square and prime number

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

• 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.

• If $p(p+2013)$ is a square number for $p$ prime then $p+2013$ has to be $p$ times a square, which means $1+\frac{2013}{p}$ has to be an integer and square. But $1+\frac{2013}{3}$, $1+\frac{2013}{11}$, and $1+\frac{2013}{61}$ are not square and for other primes $p$ this would not give an integer. – Henry Nov 15 '15 at 11:46
• The number is even, so cannot be prime – vladz Nov 15 '15 at 11:54

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 $N=671$

$$N(N+2013 ) =y^2$$ $$N^2 +2\cdot \frac{2013}{2 } \cdot N +\frac{2013^2}{4} -\frac{2013^2}{4} =y^2$$ $$4y^2 -(2N +2013)^2 =-2013^2$$ $$(2y -2N -2013 )(2y +2N +2013 ) =-2013^2$$ and you can show the above equation using the factorization of $2013^2$

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.