5

I have to show, that every number of the form $2^n$ ($n \in \mathbb{N}$) is a sum of two squares. I don't have any idea how to start here.

Any help is appreciated.

Arturo Magidin
  • 398,050
ulead86
  • 3,381
  • 3
    It's false that every number $n$ is a sum of two squares: $3$ is not a sum of two squares (and in general, no number of the form $4n+3$ is a sum of two squares; and there are others, e.g., $21$ is not a sum of two squares). Fermat's Christmas Theorem characterizes the numbers that are sums of two squares. – Arturo Magidin Jan 19 '12 at 17:53
  • Oh yes ofc, you are right, I changed it above – ulead86 Jan 19 '12 at 17:56
  • 1
    And along the way, obliterated the grammar and tag changes that had already been made to your post... – Arturo Magidin Jan 19 '12 at 17:57
  • 4
    @Daniel: A negative, but I hope helpful, comment on "I don't know how to start here." Start with $n=1$, then do $n=2$, then do $n=3$, and so on. Soon everything will be clear. Here goes: $2^1=2=1^2+1^2$; $2^2=4=2^2+0^2$; $2^3=8=2^2+2^2$; $2^4=4^2+0^2$; $2^5=4^2+4^2$; $2^6=8^2+0^2$. The pattern is obvious, and now that we know what it is, writing down a formal proof is not hard. Note that the pattern is different for $n$ odd than for $n$ even. – André Nicolas Jan 19 '12 at 18:24
  • @André: Yes you are right, I just didn't saw the pattern, nevertheless it's obvious :/ – ulead86 Jan 19 '12 at 19:02
  • @Daniel: As to the general problem of representing a number as a sum of two squares, it takes a lot more experimentation to detect patterns, and a lot of effort to prove things. But here we have for all $n$ an essentially unique representation and a simple pattern. I was trying to say that a good first step for many problems is to examine "small" cases. – André Nicolas Jan 19 '12 at 19:16

3 Answers3

11

The statement that every natural number $n$ is a sum of two squares is false. For example, $3$ is not a sum of any two squares.

Hint: Note that $2^n=2^{n-1}+2^{n-1}$ and $2^n=2^n+0$. Which statement should you use for which $n$?

Zev Chonoles
  • 129,973
6

Hint:

  1. Show that if $a$ and $b$ are each a sum of two squares, then so is $ab$.

  2. Then show that $2$ is a sum of two squares.

  3. Then use induction.

Arturo Magidin
  • 398,050
6

HINT $\rm\quad 2\ (a^2 + b^2)\ =\ (a + b)^2 + (a - b)^2\ $ yields the inductive step.

REMARK $\ $ Writing $\ 2\ = 1^2 + 1^2\ $ shows that the above is nothing but a special case of the well-known Brahmagupta–Fibonacci identity for composing sums of squares. This special doubling case was known much earlier than the general case. It plays a key role in my speculative reconstruction of $\rm\ FLT_4\equiv $ Fibonacci's Lost Theorem, the area of an integral pythagorean triangle $\rm\ne n^2\:.$

Community
  • 1
Bill Dubuque
  • 272,048