Magic Squares with Lucas and Fibonacci Numbers I am quite curious about can we construct magic squares using only Lucas and Fibonacci numbers(of course not repeating them?
If yes,  how can we construct them?
And if not , what is the proof?

Theorem: It is not possible to make a nontrivial (1x1 magic squares always work) magic square out of purely Fibonacci numbers or purely Lucas numbers. 
Proof: Assume for the sake of contradiction that we have a magic square $M$ with $n^2$ different Fibonacci numbers as elements, and $n\geq2$. Let $F_k$ be the largest entry in the magic square, and let $R$ be the row which includes $F_k$. 
First suppose either $F_k$ or $F_{k-1}$ appears in every row. Then $M$ can only be a 2x2 square. This leads to a contradiction because there exist no nontrivial 2x2 magic squares (trivial squares here mean ones consisting of all the same element). We can see this by noting the sum of any two elements in a 2x2 magic square must equal the sum of any other two elements in the square, which implies all entries are the same.
Suppose now there exists a row $H$ that neither includes $F_k$ nor $F_{k-1}$. Then 
$$\sum\limits_{x\in H}x\leq\sum\limits_{n=0}^{k-2}F_n=F_k-1<F_k<\sum\limits_{x\in R}x.$$
Hence the sum of elements in row $H$ is strictly less than the sum of elements in row $R$, so $M$ cannot be a magic square.
This completes the proof.
The case of Lucas numbers is completely analogous since as with Fibonacci numbers, we have
$$\sum\limits_{n=0}^{k-2}L_n=L_k-1,$$
which can be proven easily by induction.
In essence, both the Fibonacci numbers and the Lucas numbers are spaced too far apart from each other to form a magic square. 
