Let $F_k$ denote the $k$th Fermat number $2^{2^k}+1$. If $$A=\{F_{2n}, \ F_{2n+2}, \ F_{2n+4}, \ 4F_{2n+1}F_{2n+2}F_{2n+3}\},$$ then I want to show that for $x,y\in A$ with $x\neq y$ the term $xy+1$ is a perfect square.
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We have that $$F_M=2^{2^M}+1 \ \text{ and } \ F_N=2^{2^N}+1$$ Then $$F_MF_N+1=\left (2^{2^M}+1\right )\cdot \left (2^{2^N}+1\right )+1=2^{2^M+2^N}+2^{2^M}+2^{2^N}+1$$
Do we have to substitute the values for $M$ and $N$ to see that we get a perfect square or do we have to do something else?
You are probably talking about Fibonacci numbers.
Hint Use the equation: $F_{n+1}^2= F_{n}F_{n+2} + (-1)^n$