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Professor proposed this problem to the class today.

Suppose we had $P_1(x), P_2(x) \in \mathbb{Z[x]}$, $n, a \in \mathbb{Z}$.

How many ordered pairs exist such that $(P_1(x))^2+(P_2(x))^2=(x^n-a)^2$?

Of course, there exist trivial pairs, such as $(x^n-a,0)$, but I'm not sure where to go from here.

user231271
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1 Answers1

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Partial solution when $x^n-a$ is irreducible over $\Bbb Z[i][x]$ this means that for every $d$ divisor of $n$ $a$ is not a $d$-th power and $a$ is not of the form $-b^2$ when $n$ is even.

Answer the only possible values of $(P_1,P_2)$ are $(\mp(x^n-a),0)$ and $(0,\mp(x^n-a))$

First you can see that the degrees of $P_1$ and $P_2$ are at most $n$ because $2n=\deg((x^n-a)^2)=\deg(P_1^2(x)+P_2(x)^2)= 2\max(\deg(P_1),\deg(P_2))$, we can assume WLOG that $\deg(P_1)=n$ we have: $$(P_1+iP_2)(P_1-iP_2)=(x^n-a)(x^n-a)=(x^n-a)^2$$ And Because $\Bbb Z[i][X]$ is a unique factorization domian, Then $P_1+iP_2=c_1(x^n-a)$ and $P_1-iP_2=c_2(x^n-a)$ hence $P_2=0$ and $P_2=\mp(x^n-a)$

Elaqqad
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  • How are you deducing that $,P_1(t) = 0 = P_2(t)?\ \ $ – Bill Dubuque Apr 14 '15 at 19:52
  • The edit does not answer my question. – Bill Dubuque Apr 14 '15 at 22:52
  • Now you have a completely different proof with a different gap. Precisely how do you deduce that $,P_1 \pm i, P_2, =, c_{\pm} (x^n-a) ?\ \ $ – Bill Dubuque Apr 15 '15 at 14:39
  • I'm sure that my answer is incorrect and i'm trying to correct otherwise I will delete it – Elaqqad Apr 15 '15 at 14:50
  • I can't find a full solution, so if you have any idea you're welcome! (I conjecture that the number of solutions would be $v_2(n)$ if $a=-b^{2^{v_2(n)}}$) but I can't complete the solution – Elaqqad Apr 15 '15 at 14:54
  • @BillDubuque Using $P_a(x)=\pm x^n+\ldots$ and $\deg P_2<n$ (both fololow from thw "wlog" $\deg P_1=n$ because writing $1$ as sum of squares leaves little options), the two equtions give us$\frac{c_1+c_2}2(x^n-a)=P_1(x)=\pm x^n+\ldots\to c_1+c_2=\pm2$ and $\deg((c_1-c_2)(x^n-a))=\deg(2iP_2(x))<n$, so $c_1=c_2=\pm1$ – Hagen von Eitzen Apr 16 '15 at 06:16
  • @Elaqqad We obtain a solution from any factorization of $(x^n-a)^2$ into two factors of degree $n$. The factorizations may be of the from $f\bar g\cdot \bar fg$ with $x^n-a=fg$, but additional solutions may come from playing with equal degree factors, I guess – Hagen von Eitzen Apr 16 '15 at 06:26
  • @HagenvonEitzen yes you're right but the question is how can one find the number of divisors $x^n-a$ – Elaqqad Apr 16 '15 at 17:02