I am trying to work through Ireland and Rosen's Number theory book. Following is ex. 26, ch.8(Gauss and Jacobi sums).

Let $p$ be a prime. $p\equiv 1\mod{4}$, $\chi$ a multiplicative character of order 4 on $F_{p}$, and $\rho$ the Legendre symbol. Put $J(\chi,\rho)=a+bi$. Show

1.) $N(y^2=1-x^4)=p+\sum \rho(1-x^{4})$

2.) $2a\equiv -(-1)^{(p-1)/4}(^{2m}_{m})(p)$ where $m=(p-1)/4$

I am not a mathematician so I find it difficult many times to use abstract theory to solve concrete problems(assuming I understand the theory in the first place!). I would like to see how to solve one of these(or both) problems and at least a hint for the other one, since I am not sure how to proceed with these problems. Thank you for your time.

  • $\begingroup$ What does $(\rho)$ mean? What's $m$? I haven't read their book. $\endgroup$ – anon Feb 29 '12 at 5:26
  • $\begingroup$ @anon, sorry, that was a typo, it is fixed now. $\endgroup$ – Edison Feb 29 '12 at 5:31
  • 2
    $\begingroup$ Whose idea was it to call the Legendre symbol $\rho$ when you've already got a $p$? Yeash. $\endgroup$ – anon Feb 29 '12 at 5:34
  • $\begingroup$ Related: Explicit formula for Fermat's 4k+1 theorem $\endgroup$ – Grigory M May 8 '14 at 21:07

We need the fact that the sum of powers, $j\ge0$, $$ S(j)=\sum_{x=0}^{p-1}x^j $$ is $\equiv 0\pmod{p}$, unless $j$ is positive and divisible by $(p-1)$, and in that latter case we have $S(j)\equiv -1\pmod{p}$. This follows from the fact that when $j>0$, the non-zero terms can be reordered (according to ascending powers of a primitive root) to form a geometric sum of $p-1$ terms. Modulo $p$ such a sum is either all ones, or the sum of a full set of roots of unity of order dividing some integer $d\mid p-1$. In the latter case the sum is congruent to zero modulo $p$.

Another fact that we need is the congruence (Euler's formula) $$ \rho(x)\equiv x^{(p-1)/2}\pmod p. $$

We need to compute the residue class modulo $p$ of the character sum $$ S=\sum_{x=0}^{p-1}\rho(1-x^4)\equiv\sum_{x=0}^{p-1}(1-x^4)^{\frac{p-1}2}. $$ Here $(p-1)/2=2m$. By the binomial formula $$ (1-x^4)^{2m}=\sum_{j=0}^{2m}(-1)^j{2m\choose j}x^{4j}. $$ Summing over $x$ we get $$ S\equiv\sum_{j=0}^{2m}(-1)^j{2m\choose j}S(4j). $$ Here $S(4j)$ is non-zero modulo $p$ only, when $j= m$ or $j=2m$. Putting these pieces together we get $$ S\equiv(-1)\left((-1)^m{2m\choose m}+1\right). $$ The claim follows from this and yet another part of this exercise stating that $$ N(y^2+x^4=1)=p-1+2a. $$

The claim of that part of the exercise is derived in the same way as the other relations between Jacobi sums and the numbers $N(x^n+y^m=1)$ in this chapter, namely $$ N(y^2+x^4=1)=\sum_{i=0}^3\sum_{j=0}^1 J(\chi^i,\rho^j). $$


For a given $x$, number of solutions of $y^2=1-x^4$ is $2$ if $1-x^4$ is a square, $0$ if it isn't.

$\rho(1-x^4)$ is $1$ if $1-x^4$ is a square, $-1$ if it isn't.

So for a given $x$, $1+\rho(1-x^4)$ is the number of solutions of $y^2=1-x^4$.

Now sum on all $x$. Note: I've swept one little case under the rug, when $1-x^4=0$. This is left as an exercise for the reader.


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