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I have been doing some work on Pythagoras's Theorem with my Year 8 Maths class (7th Grade in US speak). I had them investigating what values were unobtainable for the square on the hypotenuse of right-angled triangles with integer lengths. In other words, the values that $x^2+y^2$ can take for integers $x,y$.

One student decided to start looking at the factors of $x^2+y^2$. They came up with the following question (in slightly less formal language):

Given a positive integer $d$, when is $x^2+y^2 \equiv 0\pmod d$, for integers $x$ and $y$?

I created this Geogebra resource to visualise where the zeros occur (unexpectedly creating some beautiful patterns).

Some values of $d$, like $d=17$ and $d=25$, give loads of zeros. Others, like $d=19$, give zeros only when $xy \equiv 0 \pmod {19}$.

The following table gives the number of zeros ($N$) of $x^2+y^2 \pmod d$ for integer $(x,y) \in [0,d-1]^2$, for $d=1,...,20$:

table of values

... and this scatter graph plots $N/d$ against $d$ for $d=1,...,40$:

scatter plot

I am now stuck with the following questions:

  • Given $d$, is it possible to predict $N$? (The values of $N$ seem dependent upon whether $d$ is a square number and whether $d$ occurs in a Pythagorean Triple, but with some weird exceptions like $15$ and $25$.)
  • Is there some intuitive explanation of the wavy patterns that are generated by plotting $x^2+y^2 \pmod d$, as in the Geogebra app?

Any guidance appreciated! Thank you.

pattern

$\small{\textrm{(Image: the contrast of each coordinate is proportional to $x^2+y^2 \pmod {40}$, with zeros highlighted pink.)}}$

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    $\begingroup$ It's oeis.org/A086933, maybe some referrences there can help $\endgroup$ – Sil Nov 28 '20 at 14:41
  • $\begingroup$ From the OEIS link in @Sil's comment we have $N\sim\frac{(2d+1)\pi}{8G}$. So for large $d$ the ratio $\frac Nd$ tends to $\frac\pi{4G}$. $\endgroup$ – TheSimpliFire Nov 28 '20 at 15:32
  • $\begingroup$ @TheSimpliFire It is a multiplicative function, it doesn't have an asymptotic, in contrary to its summatory function $\endgroup$ – reuns Nov 28 '20 at 18:54
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    $\begingroup$ I made the exact same picture some time ago, but for larger $d$ (and a bit more artistic freedom of the coloring): imgur.com/L9wLHrI . To get you a bit more started, the fact that $N(d)$ is multiplicative follows from the Chinese Remainder Theorem. $\endgroup$ – Kenneth Goodenough Nov 28 '20 at 23:34
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Some information copied over from oeis.org/A086933, but in the OP's notation.

$N(d)$ is a multiplicative function. That is, if $c, d$ are relatively prime, then $N(cd) = N(c)N(d)$. Also

  • $N(2^e) = 2^e$
  • $N(p^e) = ((p-1)e+p)p^{e-1}$ for primes $p \equiv 1 \mod 4$
  • $N(p^e) = p^{e-(e \bmod 2)}$ for primes $p \equiv 3 \mod 4$.

For example, $300 = 2^2\cdot3\cdot5^2$. Now, $3 \equiv 3\mod 4$ and $5\equiv 1 \mod 4$, so

  • $N(2^2) = 2^2 = 4$,
  • $N(3) = N(3^1) = 3^{1-(1\bmod 2)} = 3^{1 - 1} = 3^0= 1$,
  • $N(5^2) = ((5-1)\cdot2 + 5)\cdot 5^{2-1} = 13\cdot 5 = 65$.

Therefore $N(300) = 4\cdot1\cdot 65 = 260$.

From this we can see that $N(d) = 1$ if and only if $d$ is a product of distinct primes congruent to $3\bmod 4$.

Another formula for $N(d)$ is: $$N(d) = \sum_{r|d, r \text{ odd}} (-1)^{(r-1)/2}{\phi(r)}\frac dr$$

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  • $\begingroup$ Thank you! And to @Sil for the original link to the OEIS. I'll have a play with the Chinese Remainder Theorem to figure out why the function is multiplicative. It might also give some insight into the patterns. $\endgroup$ – Malkin Nov 29 '20 at 13:24
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    $\begingroup$ You should look at the problem using Gaussian Integers. When $d, x, y$ are relatively prime, since $d \mid (x +iy)(x-iy)$, every gaussian prime divisor of $d$ must also divide only one of $x + iy$ or $x - iy$. $\endgroup$ – Paul Sinclair Nov 29 '20 at 16:14
  • $\begingroup$ I've never come across the concept of a Gaussian prime; they seem very relevant to my problem. Thanks a lot! $\endgroup$ – Malkin Nov 29 '20 at 19:44
  • $\begingroup$ I think I've gotten my head around $N(d)$ being multiplicative. Sketching out the beginnings of a proof: for primes $p_1$, $p_2$ and integers $n_1,n_2$, choose $0\leq x_i,y_i \leq p_i^{n_i}$ such that $x_i^2+y_i^2 \equiv 0 \pmod{p_i^{n_i}}$ for $i=1,2$. Then by the CRT, $\exists ! x,y$ with $0\leq x,y \leq p_1^{n_1} p_2^{n_2}$ such that $x \equiv x_i \pmod{p_i^{n_i}}$ and $y \equiv y_i \pmod{p_i^{n_i}}$. It follows that $x^2+y^2 \equiv 0 \pmod{p_1^{n_1}p_2^{n_2}}$. The uniqueness of these $x,y$ leads to $N(p_1^{n_1}p_2^{n_2})=N(p_1^{n_1})N(p_2^{n_2})$. $\endgroup$ – Malkin Nov 29 '20 at 20:04

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