I'm reading a proof and I find it very unclear. The prove contains 5 steps, and I understand each one of them. However, author didn't say how those 5 steps really justifies the theorem. Could you please have a look at these steps and give me a hand with explaining how the thesis was proven ?


Natural number $n$ can be represented as sum of squares of two numbers iff in it's prime factorization, all prime divisors of the form $p = 4m+3$ are raised to even powers.

5 steps prove

Let's call number $n$ representable if and only if $$\exists_{x_0, y_0 \in \mathbb{N}_0}: n = x_0^2+y_0^2$$

Step 1

Observe that $1, 2$ and $p = 4m+1$ are representable

Step 2

If $n_1, n_2$ are representable then $n_1 \cdot n_2$ is representable.

Step 3

If $n$ is representable then $nz^2, z \in \mathbb{N}$ is representable

Step 4

If $p = 4m+3$ divides representable number $n = x^2+y^2$ then $p$ divides $x$ and $y$. Also $p^2 \mid n$

Step 5

If $n$ is representable and $p = 4m+3$ divides $n$ then $p^2 \mid n$ and $\frac{n}{p^2}$ is representable.

Could you please give me a hand with saying how these 5 steps actually prove thesis ?

  • $\begingroup$ Step 3 looks wrong $\endgroup$ – Hagen von Eitzen Mar 27 at 17:14
  • $\begingroup$ Yes - you were right. I updated my question. Thank you ! $\endgroup$ – Lucian Mar 27 at 17:19
  • 3
    $\begingroup$ As a side grammar note, "prove" is a verb; "proof" is the noun you intend, I think. $\endgroup$ – Brian Tung Mar 27 at 17:40
  • $\begingroup$ Thank you very much ;)) I very often mixed those two nonintentionally $\endgroup$ – Lucian Mar 27 at 17:44


Let $A$ be the set of prime divisors of $n$ of the form $4m+3$, and let $B$ be the remaining set of prime divisiors, s.t. we can rewrite $n$'s prime factorization as: $$n = \prod_{a_i \in A} a_i^{e_i} \prod_{b_i \in B} b_i^{f_i} = \alpha \beta$$ where $ \alpha = \prod_{a_i \in A} a_i^{e_i}$ and $\beta = \prod_{b_i \in B} b_i^{f_i}$.

Question states that all $e_i$ are even i.e. $\alpha$ is a perfect square. Then by step 3 (and using that the number $1$ is representable from step 1), we have $\alpha$ is representable. Also prime divisors of $\beta$ will be of the form $4m + 1$ and/or $2$. Again by step 1 and step2, $\beta$ is representable. Then by step 2, $n = \alpha \beta$ is representable.

$\Rightarrow$: Let $p$ be a prime factor of the form $4m+3$ which divides $n$. Lets' use the method of contradiction i.e. assume $p$'s exponent in prime factorization of $n$ is $2k + 1$. Then using step 5 repeatedly, we get that $\frac{n}{p^2}, \frac{n}{p^4}$ and finally $\frac{n}{p^{2k}}$ is representable. But then $p$ still divides $\frac{n}{p^{2k}}$, which by step 4 means that $p^2$ divides $\frac{n}{p^{2k}}$. But this means that $p's$ exponent in prime factorization of $n$ is atleast $2k+2$. This is a contradiction.

  • $\begingroup$ Wow! That's pretty clever! Thank you for your time! $\endgroup$ – Lucian Mar 28 at 13:16

A. If $n$ is of the stated form, then it can be represented.

Proof. (Strong) induction on $n$: Let $n$ be of the stated form and we know that every $n'<n$ of the stated form is representable. If $n$ has no prime factor at all, then $n=1$, which is representable according to step 1. So we may assume that $n$ has some prime factor $p$.

  • If $p\equiv 3\pmod 4$, then in fact $p^2\mid n$ and so by step 3, $n'=\frac{n}{p^2}$ is a number of the stated form, hence is representable by induction hypothesis, and by step 3, so is $n$

  • If $p=2$ or $p\equiv 1\pmod 4$, then $p$ is representable and $n'=\frac np$ is of the stated form, hence representable, hence by steos 1 and 2, so is $n$. $\square$

B. If $n$ is representable, then $n$ is of the stated form.

Proof. (Strong) induction again: Let $n$ be representable and we know that every representable $n'<n$ is of the stated form.

If there is no $p\equiv 3\pmod n$ with $p\mid n$, then $n$ is of the stated form and we are done. So assume $p\mid n$ for some prime $p\equiv 3\pmod 4$. Then by step 5, $p^2\mid n$ and $n'=\frac n{p^2}$ is representable. By induction hypothesis, $n'$ has the stated form, but then so has $n=p^2n'$. $\square$


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