when $F_n^2+F_m^2$ is a square for fibonacci numbers

This is a curiosity question

I'm trying to solve a Diophantine equation and I need some results about fibonnacci numbers, I encountered this problem:

For which couple of integers $(n,m)$ the number $F_n^2+F_m^2$ is a square where $F_k$ denotes the $k$th Fibonacci number?

for example when $n$ and $m$ have opposite parity we have : $$F_n^2+F_m^2=F_{n-m}F_{n+m}$$ and if we suppose $\gcd(F_n,F_m)=1$ we can conclude that $F_{n-m},F_{n+m}$ are squares, and we know that the only squares among Fibonacci numbers are $F_0=0,F_1=F_2=1$ and $F_{12}=144$

Fortunately there is a lot of proprieties related to Fibonacci numbers, but at the same time it's very difficult to find the appropriate method.

Question : Is there an efficient method to solve this problem?

• I believe this might be of some assistance. Feb 15, 2015 at 22:57
• Thank you for your answer,but his answer restricts to primitive Pythagorean triples, and according to the paper I suspect that there will not be an answer for this problem soon, because his method uses a lot the fact $gcd(F_n,F_m)=1$ which is avoided here. What I'm looking for here is a method which I can use for other similar sequences. Feb 16, 2015 at 11:26
• Just some considerations: if $a^2+b^2=c^2$, at least one number between $a$ and $b$ is a multiple of four and at least one number between $a$ and $b$ is a multiple of three, hence if $F_n^2+F_m^2$ is a square at least one number between $n$ and $m$ is a multiple of six and at least one number between $n$ and $m$ is a multiple of four. Feb 26, 2015 at 16:30
• The only parameter which can make a relation between two numbers in sequence is golden ratio c. We can use $F_n=F_m c^{n-m}$ ⇒ $F_n^2+F_m^2=(1+c^{2(n-m)}).F_m^2$.Now take m and $F_m$ as known you get an equation for $F_n$ and $n$.Letting n an integer and solving this equation gives you a rational number for $F_n$ which must be rounded to an integer closed to a number of sequence and test the result.You can also work on condition for $1+c^{2(n-m)}=k^2$i.e for which values of m and n, $(1+c^{2(n-m)})$ can be perfect square. Feb 17, 2019 at 5:13