Let $a^{2} = n+27$ and $b^{2} = n-62$. Taking the difference:  $a^{2}-b^{2} = (a+b)(a-b) = n+27-(n-62) = 89$  Since $89$ is prime, that means $a+b=89$ and $a-b=1$, which means $a=45$ and $b=44$.  That means $n=a^{2}-27 = 45^{2}-27 = 2025-27=1998$. 
What is the greatest natural number $$n$$ such that $$n+27$$ and $$n-62$$ are squares of natural number? 
A. $$598$$ 
B. $$1598$$ 
C. $$3998$$ 
D. $$1998$$ 
E. Such a number does not exist 