7
$\begingroup$

Let $F_n=2^{2^n}+1$ be the Fermat number. How to represent the Fermat number $F_n$ for $n \geq 3$ as a sum of three squares of different natural numbers? For example for $n=3$ we have $$ F_3=257=5^2+6^2+14^2. $$ Is there any simple procedure to write out such representations for another $n$?

$\endgroup$
  • 4
    $\begingroup$ $$F_n= (2^{2^{n-1}})^2+1^2 +0^2$$ $\endgroup$ – N. S. Sep 17 '14 at 16:01
  • $\begingroup$ I am sorry, the numbers must be natural and different. Zero is not allowed. I have edited the question. $\endgroup$ – Leox Sep 17 '14 at 16:04
  • 1
    $\begingroup$ Does this question have some motivation behind it? Is it from a competition? Or just a general curiosity? $\endgroup$ – Alex R. Sep 17 '14 at 16:45
  • 2
    $\begingroup$ Another trivial case is $x=y=2^{2^{n-2}},z=2^{2^{n-1}}-1$... Obviously violates the "distinct numbers" rule. $\endgroup$ – abiessu Sep 17 '14 at 18:29
12
+100
$\begingroup$

Let $X=2^{2^{n-1}}$. Then $X$ is congruent to $1$ modulo $3$, so

$$ F_n=X^2+1=\bigg(\frac{2X+1}{3}\bigg)^2+ \bigg(\frac{2X-2}{3}\bigg)^2+ \bigg(\frac{X+2}{3}\bigg)^2 $$

The values are all distinct, except when $X=1$ or $4$ (which are excluded since $n\geq 3$).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.