But How?

Number Theory Level 3

Does there exist a Fibonacci number $x$ such that $25x^4 + 40x^2 + 16$ is a perfect fourth power ?

Yes No

2 solutions

Aditya Narayan Sharma
May 12, 2016

Proceeding simply , Given expression is $\displaystyle (5x^2+4)^2$ , Since it's a perfect $4^{th}$ power

$\displaystyle \implies 5x^2+4=k^2$

$\displaystyle \implies 5x^2=(k+2)(k-2)$

We have two possibilities ,

$\displaystyle \begin{cases} 5=k+2,x^2=k-2 \\ 5=k-2,x^2=k+2 \end{cases}$

Only the second case yields a solution $x=3$ which is $F_3$

@Aditya Narayan Sharma Your method is wrong!!! Notice that x=1 is also a solution........The error occurs when you said that "We have two possibilites".......That won't always work!!!!

Aaghaz Mahajan - 2 years, 9 months ago

Agreed, in fact all even Fibonacci numbers are solutions.

Alex Burgess - 2 years, 4 months ago

By this I mean $F_n$ for even $n$ , such as $1, 3, 8, 21 ...$ etc.

Alex Burgess - 2 years, 4 months ago

Your comment is partial and wrong, any method which is legitimate and produces an answer is correct (though partially) . So your word 'Your method is wrong' is incorrect and you're weakened by your logic. Secondly, it asks for does there exist, if I could pull an example from thin air out of the blue then that would count as a solution too. Take an example, does there exists integers $a,b,c$ such that $a^2+b^2=c^2$ and I'll say yes - $(a,b,c)=(3,4,5)$ is one such. That's it. I never aimed at giving you a full proof solution. That's unnecessary and preposterous , work on your comments they're deluding,misleading and deceptive. You did extra work for a question, the below given solution is not the ideal one for this question and rightly the author of the other answer has pointed out the expected question.

Aditya Narayan Sharma - 2 years, 4 months ago

Ok, so Aaghaz's comment is very direct and the exclamation marks are unnecessary, and is also partially flawed, as the 2 possibilities you arrived that includes $x=1$ as a solution.

That being said, some of the logic is flawed in your answer. The line "We have two possibilities" is incorrect, as there are infinite many solutions to the equation above it. Unfortunately you can't just factor out the 5 from $5 x^2$ in this way:

Note that if $x=abc, (k \pm 2)=5 a^2 b , (k \mp 2)=b c^2$ is valid. Eg $x=6, 5x^2 = (5 * 2^2) * (3^2) [a=2,b=1,c=3]$ or $(5 * 6) * (6) [a=1, b=6, c=1]$ . And the last conclusion "Only the second case yields a solution" is incorrect, as the first yields $x=1$ .

I agree, if you said $x=3$ is a solution, that would be valid and logically sound. But the way you stated the proof is such that you are suggesting that "The only solution is $x=3$ from logic $A, B$ and $C$ ", which is incorrect.

I did extra work for the question because I found the question could have been better to demonstrate the property I described. The solution I posted is clearly not optimal for the current question, as at the moment I can come along, test $x=1$ , get $81$ , and go yes, and move on. But there is more beauty to behold in the question, hence I commented to show that.

Alex Burgess - 2 years, 4 months ago

I'm not meaning to attack your answer, as your solution is rather neat, but merely it would be more logically sound if it said "There may be countless possibilities, but 1 case occurs when: " ... " leading to 2 solutions $x=1$ and $x=3$ , hence the answer is Yes.".

Alex Burgess - 2 years, 4 months ago

Alex Burgess
Jan 31, 2019

I think the question is flawed by in not asking the true question:

"Is it true that $25x^4 \pm 40x^2 + 16$ is a perfect fourth power iff $x$ is a Fibonacci number?"

This breaks down to: $5x^2\pm4 = k^2$ , or $k^2-5x^2 = \mp 4$ .

Note that for the $n^{th}$ Fibonacci number $F_n$ : $F_n = \frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}$ .

Hence $5F_n^2 = \phi^{2n}-2(-1)^n+(\phi)^{-2n} = \phi^{2n}+2(-1)^n+(-\phi)^{-2n} -4(-1)^n = (\phi^{n}+(-\phi)^{-n})^2-4(-1)^n$

So $L_n^2-5F_n^2=4(-1)^n$

Where $L_n = \phi^{n}+(-\phi)^{-n}$ is the $n^{th}$ Lucas number.

This proves that for every $n, 25F_n^4 + (-1)^n 40F_n^2 + 16 = L_n^4$ , so, for every even $n, 25F_n^4 + 40F_n^2 + 16 = L_n^4$ .

These can be proved to be the only solutions using some number fields in $\mathbb{Z} [\sqrt{5}]$

Excellent!!!!! This was the same method I followed while solving the question!!!

Aaghaz Mahajan - 2 years, 4 months ago

But How?

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