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 2016-12-11, 11:30 #1 Nick     Dec 2012 The Netherlands 1,453 Posts Basic Number Theory 12: sums of two squares Which positive integers can be written as the sum of 2 squares (i.e. in the form $$a^2+b^2$$ where $$a,b$$ are also integers)? This time, we use what we have learned about the Gaussian integers to answer this question about the rational (ordinary) integers. For any real (or complex) numbers $$a,b$$ with $$a\neq 0$$, there is only one number $$x$$ satisfying the equation $$ax+b=0$$. For any numbers $$a,b,c$$ with $$a\neq 0$$, there are at most 2 distinct values of $$x$$ for which $$ax^2+bx+c=0$$. (You may know a formula for them.) For any numbers $$a,b,c,d$$ with $$a\neq 0$$, there are at most 3 distinct values of $$x$$ such that $$ax^3+bx^2+cx+d=0$$, and so on. This holds whether we work in $$\mathbb{Z},\mathbb{Z}[i],\mathbb{Q}, \mathbb{R}$$ or $$\mathbb{C}$$. It also works in the integers modulo a prime number, but it need not be true in the integers modulo $$n$$ if $$n>1$$ but not prime. For example, the equation $$x^2-\bar{1}=\bar{0}$$ has 4 distinct solutions in the integers modulo 8: $$\bar{1},\bar{3},\bar{5}$$ and $$\bar{7}$$ (or equivalently $$\pm\bar{1},\pm\bar{3}$$). This stems from the fact that there are zero divisors in the integers modulo 8, e.g. $$\bar{2}\times\bar{4}=\bar{0}$$. Proposition 60 Let $$R$$ be a set of numbers including 1 which is closed under subtraction and multiplication and does not contain any zero divisors. Let $$n$$ be a positive integer, and $$a_0,a_1,a_2,\ldots,a_n$$ elements of $$R$$ with $$a_n\neq 0$$. Define $$Z=\{x\in R:a_0+a_1x+a_2x^2+\ldots +a_nx^n=0\}$$. Then $$Z$$ contains at most $$n$$ distinct elements. proof Induction on $$n$$. In the case $$n=1$$, for any $$x,y\in Z$$ we have $$a_0+a_1x=0=a_0+a_1y$$ so $$a_1(x-y)=0$$. As $$a_1$$ and $$x-y$$ are elements of $$R$$, which has no zero divisors, and $$a_1\neq 0$$, it follows that $$x-y=0$$ and therefore $$x=y$$. So all elements of $$Z$$ are equal, hence $$Z$$ contains at most 1 element. Take any integer $$N>1$$ and suppose the proposition holds for all integers $$nd^2$$, then putting $$a=|d|,b=|c|$$ is sufficient. UNIQUENESS Take any positive integers $$a,b,c,d$$ with \(a