I watched a YouTube video of an episode of Who Wants To Be A Millionaire?, in which a contestant was presented with a list of perfect squares. He was asked to choose the number that was also the sum of two smaller perfect squares. This video got me wondering if there is a way to find the correct answer choice. I thought he could maybe try each answer choice and see if there exists two integers, the sums of which add up to the chosen number.
Pythagorean triples come to mind.
Note: Definitions for certain terms are posted below.
It is quite easy if you use the ring $\mathbb{Z}[\text{i}]$ of Gaussian integers. For $w\in\mathbb{Z}[\text{i}]$, write $\bar{w}$ for the complex conjugate of $w$. Factorize $z\in \mathbb{N}$ as a product of primes $$z=2^mp_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}q_1^{s_1}q_2^{s_2}\cdots q_l^{s_l}$$ over $\mathbb{Z}$, where the $p_i$'s and the $q_j$'s are positive odd primes such that $p_1,p_2,\ldots,p_k\equiv 1\pmod{4}$ and $q_1,q_2,\ldots,q_l\equiv -1\pmod{4}$. It is well known that $z$ is a sum of two square if and only if each of the $s_j$'s is even. From now on, we assume that $s_j$ is even for every $j=1,2,\ldots,l$.
For $i=1,2,\ldots,k$, $p_i$ can be factorized into the product $\pi_i\bar{\pi}_i$, where $\pi_i\in\mathbb{Z}[\text{i}]$ is a Gaussian prime. Also, $2=(1+\text{i})(1-\text{i})$. Hence, you can pick $\mu_i\in\left\{0,1,\ldots,r_i\right\}$ for each $i=1,2,\ldots,k$ and write $$z=\left((1+\text{i})^m\,\prod_{i=1}^k \pi_i^{\mu_i}\bar{\pi}_i^{r_i-\mu_i}\,\prod_{j=1}^l q_j^{s_j/2}\right)\left((1-\text{i})^m\,\prod_{i=1}^k \bar{\pi}_i^{\mu_i}\pi_i^{r_i-\mu_i}\,\prod_{j=1}^l q_j^{s_j/2}\right)\,.$$ Hence, if $x,y\in\mathbb{Z}$ are such that $$x+y\text{i}=u\,(1+\text{i})^m\,\prod_{i=1}^k \pi_i^{\mu_i}\bar{\pi}_i^{r_i-\mu_i}\,\prod_{j=1}^l q_j^{s_j/2}\,,$$ where $u\in\{-1,+1,-\text{i},+\text{i}\}$, then $x-y\text{i}=\bar{u}\,(1-\text{i})^m\,\prod_{i=1}^k \bar{\pi}_i^{\mu_i}{\pi}_i^{r_i-\mu_i}\,\prod_{j=1}^l q_j^{s_j/2}$ and $$z=(x+y\text{i})(x-y\text{i})=x^2+y^2\,.$$
It can be shown that there are $4\prod_{i=1}^k\left(r_i+1\right)$ ways to find $(x,y)\in\mathbb{Z}^2$ such that $x^2+y^2=z$. Since there are $4$ ways to choose $u$ and, for $i=1,2,\ldots,k$, there are $r_i+1$ ways to choose $\mu_i$, we have found all possible pairs $(x,y)\in\mathbb{Z}^2$ such that $x^2+y^2=z$. If signs and ordering of $x,y$ are ignored, there are exactly $\left\lceil\frac{1}{2}\prod_{i=1}^k\left(r_i+1\right)\right\rceil$ possible pairs. If you want to exclude the case where $x$ or $y$ may be zero, then there are two cases:
(1) All $r_i$'s are even: there are $\frac12\left(\prod_{i=1}^k\left(r_i+1\right)-(-1)^m\right)$ pairs of nonzero $x,y$, up to signs and order;
(2) At least one of the $r_i$'s is odd: there are $\frac{1}{2}\prod_{i=1}^k\left(r_i+1\right)$ pairs of nonzero $x,y$, up to signs and order.
For example, let $z:=90=2\cdot 5\cdot 3^2$, with $2=(1+\text{i})(1-\text{i})$ and $5=(2+\text{i})(2-\text{i})$. Hence, we can take $x,y\in\mathbb{Z}$ such that $x+y\text{i}=(1+\text{i})\cdot (2+\text{i})\cdot 3=3+9\text{i}$, or $(x,y)=(3,9)$. In fact, there are $8$ ways to write $z=90$ as a sum of two squares; i.e., $(\pm3,\pm9)$ and $(\pm9,\pm3)$ are all possible values of $(x,y)$ (only one pair---$(x,y)=(3,9)$---if sings and ordering are ignored). Another example is $z:=50$. There are $12$ ways to write $z$ as a sum of two squares; i.e., $(\pm5,\pm5)$, $(\pm1,\pm7)$, and $(\pm7,\pm1)$ are all possible values of $(x,y)$ (only two pairs---$(x,y)=(5,5)$ and $(x,y)=(1,7)$---if signs and ordering are ignored).
In the case where $z=t^2$ for some $t\in\mathbb{N}$, the situation is not changed at all. You still need to factorize $z$ as above. For example, if $z=13^2\cdot 17^2\cdot 19^2$, you notice that $13=(3+2\text{i})(3-2\text{i})$ and $17=(4+\text{i})(4-\text{i})$. So, you can take $x,y\in\mathbb{Z}$ such that $x+y\text{i}=(3+2\text{i})\cdot(4+\text{i})\cdot 19=19(10+11\text{i})$, or $(x,y)=(10\cdot 19,11\cdot 19)$. Indeed, there are $36$ possible solutions $(x,y)\in\mathbb{Z}^2$ to $x^2+y^2=z$, and, up to signs and order, there are $5$ of them: $(10\cdot 19,11\cdot 19)$, $(5\cdot 19,14\cdot 19)$, $(1\cdot 13\cdot 19,4\cdot 13\cdot 19)$, $(2\cdot 17\cdot 19,3\cdot 17\cdot 19)$, and $(0,13\cdot 17\cdot 19)$.
P.S. If you are asking how to factorize an integer prime $p\in\mathbb{N}$ such that $p\equiv 1\pmod{4}$ into a product $\pi\bar{\pi}$ of conjugate Gaussian primes $\pi$ and $\bar{\pi}$, then you first need to solve for $t^2\equiv -1\pmod{p}$ (you can, in fact, take $t:=\left(\frac{p-1}{2}\right)!$, but it is probably easier to search for $t\in\left\{1,2,\ldots,\frac{p-1}{2}\right\}$ directly, when $p$ is large). The ring $\mathbb{Z}[\text{i}]$ has a division algorithm. Therefore, you can define a Euclidean algorithm in $\mathbb{Z}[\text{i}]$. Then, you take $\pi$ to be a greatest common divisor of $p$ and $t+\text{i}$.
For example, if $p=29$, we take $t:=12 \equiv \left(\frac{29-1}{2}\right)!\pmod{29}$. Perform the Euclidean algorithm to get $p=(t+\text{i})\cdot 2+(5-2\text{i})$ and $t+\text{i}=(5-2\text{i})\cdot(2+\text{i})+0$, from which we conclude that $5-2\text{i}$ is a greatest common divisor of $p$ and $t+\text{i}$. Hence, $\pi$ can be taken to be $5-2\text{i}$. Note that $p=\pi\bar{\pi}$, as desired. For more information, see the notes by Keith Conrad below.
Definitions
Factorize: "resolve or be resolvable into factors." Google
Gaussian integer: "a complex number whose real and imaginary parts are both integers" Wikipedia
Reference
http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf
$18^2 + 24^2 = 324 + 576 = 900 = 30^2$
– bthmas Jun 28 '15 at 06:04