Suppose that $N$ has at least 3 divisors other than $1$ and $N$, then the sum of squares of its divisors is at least $N^2+1+d^2+(N/d)^2+\sqrt{N}^2\geq N^2+3N+1$. If $N$ has less than $2$ divisors other than $1$ and $N$, then the sum of squares of its divisors is at most $N^2+1+\sqrt{N}^2=N^2+n+1$. Then $N=pq$ for some primes $p$ and $q$, so $1+p^2+q^2+(pq)^2=(pq)^2+3pq \implies p^2+q^2+1-3pq=0$. By standard Vieta Jumping, the only $(x,y)$ satisfying $x^2+y^2+1-3xy=0$ are pairs $(a_m,a_{m+1})$ generated by $a_1=a_2=1$ and $a_{n+1}=3a_n-a_{n-1}$. $3F_{m+2}-F_m=2F_{m+2}+F_{m+1}=F_{m+2}+F_{m+3}=F_{m+4}$, so by induction, $a_n=F_{2n-3}$ for $n\geq 2$, so $\{p,q\}=\{F_{2n-1},F_{2n+1}\}$ for some $n$, as desired.

Basically the same as above. The key idea is the following claim. Claim: $N$ has at most four divisors. Proof: Let $1, d, \tfrac{N}{d}, N, e$ be five different divisors of $N$ such that $e > \sqrt{N}$. Then $$N(N+3) \geq 1+\left(d^2+\frac{N^2}{d^2}\right) + N^2 + e^2 \geq 1 + 2N + N^2 + N > N(N+3)$$which is a contradiction. If $n = p^t$ is a prime power, then $1 + p^2 + p^4 + \hdots + p^{2t} = p^t(p^t+3)$ which is not possible as LHS is not divisible by $p$. Thus the only case left is when $n = pq$ is semiprime, which the equation is equivalent to $p^2+q^2+1 = 3pq$. This is a standard Vieta Jumping exercise and we outline the proof below. Lemma: All pairs $(a,b)$ of positive integers satisfying $a\geq b$ and $a^2+b^2+1=3ab$ are $(1,1)$ and $(F_{2n-1},F_{2n+1})$ where $n=1,2,\hdots$. Proof: Let $T$ denote the set of all solutions. Then $(a,b)\in T$ implies $(b,3b-a)$ satisfies the equation. We claim that this pair is actually in $T$ unless $(a,b)=(1,1)$. First, $3b-a = \tfrac{b^2+1}{a} > 0$ thus $3b-a$ is actually a positive integer. Assume that $b < 3b-a$, then $a < 2b$. Thus $$a^2+b^2+1 \leq a(2b-1) + (ab) + 1 = 3ab - a +1$$thus $a=1$ which imply $b=1$. Therefore through a series of this transformation, any pair in $T$ must go to $(1,1)$. Thus all solutions are generated by the sequence $(a_n, a_{n+1})$, defined by $a_1 = a_2 = 1$ and $a_{n+1} = 3a_n-a_{n-1}$. A simple induction verifies that this actually matched with the solution sets describe above.

Very nice problem!

Almost the same as above ones, but with a different finish (Although I use a "well known" fact) First, easily check that $N = p$, $N = p^2$ fail. Now, let $d$ be the smallest divisor of $n$ with $\sqrt{N} > d > 1$ Note that $d^2 + \frac{N^2}{d^2} > 2N$ by AM-GM (with equality not holding), so at most one such divisor exists, if $z = \sqrt{N}$ is another divisor, it still contributes $N$ so the final sum of squares is $> N^2 + 3N+1$, impossible. So, we may let $N = pq$ for primes $p,q$ Now, note that we need $1^2 + p^2 + \frac{N^2}{p^2} + N^2 = N^2 + 3N$. Writing this is a quadratic in $N$, we need the discriminant, $5p^4 - 4p^2 = p^2(5p^2-4)$ to be a perfect square, which needs $5p^2 - 4$ to be a perfect square. But its "well known" that this happens only when $p$ is a Fibbonacci number. Repeating the same thing, we see that $q$ is also a Fibbonacci number. So, $N = pq$ can be written as the product of two Fibbonacci numbers, as desired. $\blacksquare$

For this condition to be satisfied, $N$ must have only 2 prime divisors. $N = pq$.$5p^2-4$ and $5q^2-4$ are perfect squares under the condition.So $p$ and $q$ are Fibonacci numbers.

https://en.wikipedia.org/wiki/Pell%27s_equation