[Math] Perfect square and prime number

elementary-number-theorynumber theory

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.

Best Answer

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 $N=671$

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