a) If $n$ is prime, then $ n(n+2013) $ must be a multiple of $n^2$, so that $n+2013$ is a multiple of $n$. Hence the prime $n$ divides $2013$, which leaves the three cases $n=3$, $n=11$, $n=61$ and none of these yields a square. b) If we try $n=m^2$, then $m^2+2013$ must be square. Since the difference between two consecutive squares $m^2$ and $(m+1)^2$ equals $2m+1$, we try $2m+1=2013$ and get $m=1006$. Hence $n=1006^2$ works.

there are three more pairs other than $n$$={1006}^{2}$.We see that $gcd(n,n+2013)=1$ by the solution of (a).Therefore if $n(n+2013)={x}^{2}$ for some $x$ then $n=y^{2}$ and $n+2013={z}^{2}$ such that $yz=x$.Therefore $2013=(z-y)(z+y)$.Now put $z+y=61$ and$z-y=33$.Solving we get $y=14$ i.e. $n=y^{2} ={14}^{2}=196$.Similarly we can do for other combinations.So we get the solutions of $n$ as $n=({14}^{2},{1006}^{2},{334}^{2},{86}^{2})$

The complete list is actually $196, 671, 976, 1875, 4575, 7396, 9251, 15616, 29700, 91091, 111556, 336675, 1012036$. You only included $196, 1012036, 111556, 7396$. The correct solution is to let $k^2=n(n+2013)$. Then multiply by 4 and complete the square to get $(2n+2013)^2-4k^2=2013^2$. Then we can factor the left side as $(2n+2013+2k)(2n+2013-2k)=2013^2$. So we take pairs of factors of $2013^2$, say $ab=2013^2$, and take their sum, which equals $4n+4026$. We then subtract $4026$ from this sum and divide by 4. The list I gave above is the result of doing this for all factors of $2013^2$ except for $2013$ itself. The problem with your solution is that you just assume that $n=y^2$ and $n+2013=z^2$, when in reality, you could have $n=ky^2$ and $n+2013=kz^2$ for some squarefree number $k$. That's what happens in the case of $n=671$.

n^2+2013n-a^2=0 D= 2013^2+4a^2=b^2 (b-2a)(b+2a)=3^2×11^2×61^2 From here it is easy

Do not complicate. Math_science has simple solution. Let $n(n+2003)=k^2$. Multiplying this by $4$ we get $(2n+2003)^2-2003^2=(2k)^2$. Now take difference of squares to left moving $2003^2$ to the right and then play with several cases.

This is very easy. First if $ n(n+2013)=k^2 $ if we assume $ n $ is prime we get that $ n|k \implies n^2|k^2 \implies n^2|n(n+2013) \implies n|n+2013 \implies n|2013$ which is now obvious contradiction. Now we have $ n(n+2013)=k^2 \*4$ equivalent $4*n^2 +4*2013 = 4*k^2$ equivalent $(2*n+2013)^2 -4*k^2=2013^2$ equivalent $(2*n+2013 -2*k)*(2*n+2013 +2*k)=2013^2 $ which is now obvious .