# Find number of ordered pairs $(a, b)$ satisfying $a^2+b^2=2^3\cdot3^4\cdot5\cdot17^2$

Find number of ordered pairs $$(a, b)$$ satisfying $$a^2+b^2=2^3\cdot3^4\cdot5\cdot17^2$$. By rearranging the terms, I found a pair (918, 306). But I wonder if there is a systematic way to solve for the number of pairs? Any hint will be appreciated!

• Mathworld gives a formula for the number of such pairs. Commented Jun 15, 2020 at 4:42

Since the ring of Gaussian integers $$\mathbb{Z}[i]$$ is a Euclidean domain, hence a UFD,

$$r_2(n) = \left|\{(a,b)\in\mathbb{Z}\times\mathbb{Z}: a^2+b^2 = n\}\right|$$ is a constant multiple of a multiplicative function, namely $$r_2(n) = 4(\chi_4*1)(n)=4\sum_{d\mid n}\chi_4(d)\quad\text{where}\quad \chi_4(d)=\left\{\begin{array}{rcl}0&\text{if}& d\equiv 0\pmod{2}\\ 1&\text{if}&d\equiv 1\pmod{4}\\ -1&\text{if}& d\equiv -1\pmod{4}\end{array}\right.$$ In particular $$r_2(n)$$ only depends on the prime factors & exponents in the factorization of $$n$$:

$$r_2(2^3\cdot 3^4\cdot 5\cdot 17^2)= r_2(3^4\cdot 5\cdot 17^2)=r_2(5\cdot 17^2)=4\tau(5\cdot 17^2)=4\cdot 2\cdot 3=24.$$

Here the possible values for $$a$$ and $$b$$, up to sign: $$162,306,\color{red}{666},702,918,954$$
and a practical algorithm for computing $$r_2(n)$$ by hand, given the factorization of $$n$$:

• replace $$n$$ with $$n/2^{\nu_2(n)}$$, i.e. drop the eventual factor $$2^\alpha$$
• if some prime $$\equiv -1\pmod{4}$$ appears with an odd exponent, return 0. Otherwise, drop all these factors
• return four times the number of divisors, i.e. increase all the exponents in the factorization by one and return four times their product.

Related: there are circles which go through an arbitrarily high number of lattice points, since $$r_2(5^k)=4k+4$$.

• Nice, +1!${}{}{}$ Commented Jun 15, 2020 at 10:22

Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica code:

In[1]:=Length[FullSimplify[
Solve[{a^2 + b^2 == 2^3*3^4*5*17^2, b > a}, {a, b}, Integers]]]


Running the code gives:

Out[1]=12


Looking for the solutions, we can see:

In[2]:=FullSimplify[
Solve[{a^2 + b^2 == 2^3*3^4*5*17^2, b > a}, {a, b}, Integers]]

Out[2]={{a -> -954, b -> -162}, {a -> -954, b -> 162}, {a -> -918,
b -> -306}, {a -> -918, b -> 306}, {a -> -702,
b -> -666}, {a -> -702, b -> 666}, {a -> -666,
b -> 702}, {a -> -306, b -> 918}, {a -> -162, b -> 954}, {a -> 162,
b -> 954}, {a -> 306, b -> 918}, {a -> 666, b -> 702}}


So, we can see that when we have $$(\text{a},\text{b})$$ where $$\text{b}>\text{a}$$ there are $$12$$ solutions to that problem.