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Yes, this is standard. More generally, we have the following

Theorem. Let $p$ be an odd prime, and let $a_1,\dots,a_k\in\mathbb{F}_p^\times$. Then the number of solutions of the quation $a_1x_1^2+\cdots +a_kx_k^2=1$ in $\mathbb{F}_p$ equals \begin{align*} p^{k-1}-\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 0\pmod{4};\\ p^{k-1}+\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 1\pmod{4};\\ p^{k-1}-\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 2\pmod{4};\\ p^{k-1}+\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 3\pmod{4}. \end{align*}

P.S. I am sure there is a reference for this, but I found it easier to copy from my notes.

Added 1. As Gerry Myerson pointed out (in a comment he deleted), the special case $a_1=\dots=a_k=1$ is Proposition 8.6.1 in Ireland-Rosen: A classical introduction to modern number theory (2nd edition).

Added 2. For the sake of completeness, I provide the proof. Let $e_p:\mathbb{F}_p\to\mathbb{C}^\times$ be the standard additive character, $\chi:\mathbb{F}_p^\times\to\{\pm 1\}$ the nontrivial quadratic character, and $$\lambda:=\begin{cases} 1,&\qquad p\equiv 1\pmod{4};\\ i,&\qquad p\equiv 3\pmod{4}.\end{cases}$$ Let us remark that $\lambda^2=\chi(-1)$. If $n$ denotes the number of solutions in the theorem, then \begin{align*} pn&=\sum_{x_1,\dots,x_k\in\mathbb{F}_p}\ \sum_{m\in\mathbb{F}_p}\ e_p(m(a_1x_1^2+\cdots +a_kx_k^2-1))\\ &=\sum_{m\in\mathbb{F}_p}e_p(-m)\sum_{x_1\in\mathbb{F}_p}e_p(ma_1x_1^2)\ \cdots\sum_{x_k\in\mathbb{F}_p}e_p(ma_kx_k^2)\\ &=p^k + \sum_{m\in\mathbb{F}_p^\times}e_p(-m)\bigl\{\chi(ma_1)\lambda\sqrt{p}\bigr\}\cdots \bigl\{\chi(ma_k)\lambda\sqrt{p}\bigr\}\\ &=p^k + \chi(a_1\dots a_k)\lambda^k p^{\frac{k}{2}}\sum_{m\in\mathbb{F}_p^\times}e_p(-m)\chi(m)^k. \end{align*} The inner sum equals $-1$ or $\chi(-1)\lambda\sqrt{p}$ depending on whether $k$ is even or odd, therefore $$n=\begin{cases} p^{k-1}-\chi((-1)^{\frac{k}{2}}a_1\dots a_k)p^{\frac{k-2}{2}},&k\equiv 0\pmod{2};\\ p^{k-1}+\chi((-1)^{\frac{k-1}{2}}a_1\dots a_k)p^{\frac{k-1}{2}},&k\equiv 1\pmod{2}. \end{cases}$$

Yes, this is standard. More generally, we have the following

Theorem. Let $p$ be an odd prime, and let $a_1,\dots,a_k\in\mathbb{F}_p^\times$. Then the number of solutions of the equation $a_1x_1^2+\cdots +a_kx_k^2=1$ in $\mathbb{F}_p$ equals \begin{align*} p^{k-1}-\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 0\pmod{4};\\ p^{k-1}+\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 1\pmod{4};\\ p^{k-1}-\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 2\pmod{4};\\ p^{k-1}+\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 3\pmod{4}. \end{align*}

P.S. I am sure there is a reference for this, but I found it easier to copy from my notes.

Added 1. As Gerry Myerson pointed out (in a comment he deleted), the special case $a_1=\dots=a_k=1$ is Proposition 8.6.1 in Ireland-Rosen: A classical introduction to modern number theory (2nd edition).

Added 2. For the sake of completeness, I provide the proof. Let $e_p:\mathbb{F}_p\to\mathbb{C}^\times$ be the standard additive character, $\chi:\mathbb{F}_p^\times\to\{\pm 1\}$ the nontrivial quadratic character, and $$\lambda:=\begin{cases} 1,&\qquad p\equiv 1\pmod{4};\\ i,&\qquad p\equiv 3\pmod{4}.\end{cases}$$ Let us remark that $\lambda^2=\chi(-1)$. If $n$ denotes the number of solutions in the theorem, then \begin{align*} pn&=\sum_{x_1,\dots,x_k\in\mathbb{F}_p}\ \sum_{m\in\mathbb{F}_p}\ e_p(m(a_1x_1^2+\cdots +a_kx_k^2-1))\\ &=\sum_{m\in\mathbb{F}_p}e_p(-m)\sum_{x_1\in\mathbb{F}_p}e_p(ma_1x_1^2)\ \cdots\sum_{x_k\in\mathbb{F}_p}e_p(ma_kx_k^2)\\ &=p^k + \sum_{m\in\mathbb{F}_p^\times}e_p(-m)\bigl\{\chi(ma_1)\lambda\sqrt{p}\bigr\}\cdots \bigl\{\chi(ma_k)\lambda\sqrt{p}\bigr\}\\ &=p^k + \chi(a_1\dots a_k)\lambda^k p^{\frac{k}{2}}\sum_{m\in\mathbb{F}_p^\times}e_p(-m)\chi(m)^k. \end{align*} The inner sum equals $-1$ or $\chi(-1)\lambda\sqrt{p}$ depending on whether $k$ is even or odd, therefore $$n=\begin{cases} p^{k-1}-\chi((-1)^{\frac{k}{2}}a_1\dots a_k)p^{\frac{k-2}{2}},&k\equiv 0\pmod{2};\\ p^{k-1}+\chi((-1)^{\frac{k-1}{2}}a_1\dots a_k)p^{\frac{k-1}{2}},&k\equiv 1\pmod{2}. \end{cases}$$

Yes, this is standard. More generally, we have the following

Theorem. Let $p$ be an odd prime, and let $a_1,\dots,a_k\in\mathbb{F}_p^\times$. Then the number of solutions of the quation $a_1x_1^2+\cdots +a_kx_k^2=1$ in $\mathbb{F}_p$ equals \begin{align*} p^{k-1}-\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 0\pmod{4};\\ p^{k-1}+\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 1\pmod{4};\\ p^{k-1}-\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 2\pmod{4};\\ p^{k-1}+\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 3\pmod{4}. \end{align*}

P.S. I am sure there is a reference for this, but I found it easier to copy from my notes.

Added. As Gerry Myerson pointed out (in a comment he deleted), the special case $a_1=\dots=a_k=1$ is Proposition 8.6.1 in Ireland-Rosen: A classical introduction to modern number theory (2nd edition).

Yes, this is standard. More generally, we have the following

Theorem. Let $p$ be an odd prime, and let $a_1,\dots,a_k\in\mathbb{F}_p^\times$. Then the number of solutions of the quation $a_1x_1^2+\cdots +a_kx_k^2=1$ in $\mathbb{F}_p$ equals \begin{align*} p^{k-1}-\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 0\pmod{4};\\ p^{k-1}+\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 1\pmod{4};\\ p^{k-1}-\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 2\pmod{4};\\ p^{k-1}+\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 3\pmod{4}. \end{align*}

P.S. I am sure there is a reference for this, but I found it easier to copy from my notes.

Added 1. As Gerry Myerson pointed out (in a comment he deleted), the special case $a_1=\dots=a_k=1$ is Proposition 8.6.1 in Ireland-Rosen: A classical introduction to modern number theory (2nd edition).

Added 2. For the sake of completeness, I provide the proof. Let $e_p:\mathbb{F}_p\to\mathbb{C}^\times$ be the standard additive character, $\chi:\mathbb{F}_p^\times\to\{\pm 1\}$ the nontrivial quadratic character, and $$\lambda:=\begin{cases} 1,&\qquad p\equiv 1\pmod{4};\\ i,&\qquad p\equiv 3\pmod{4}.\end{cases}$$ Let us remark that $\lambda^2=\chi(-1)$. If $n$ denotes the number of solutions in the theorem, then \begin{align*} pn&=\sum_{x_1,\dots,x_k\in\mathbb{F}_p}\ \sum_{m\in\mathbb{F}_p}\ e_p(m(a_1x_1^2+\cdots +a_kx_k^2-1))\\ &=\sum_{m\in\mathbb{F}_p}e_p(-m)\sum_{x_1\in\mathbb{F}_p}e_p(ma_1x_1^2)\ \cdots\sum_{x_k\in\mathbb{F}_p}e_p(ma_kx_k^2)\\ &=p^k + \sum_{m\in\mathbb{F}_p^\times}e_p(-m)\bigl\{\chi(ma_1)\lambda\sqrt{p}\bigr\}\cdots \bigl\{\chi(ma_k)\lambda\sqrt{p}\bigr\}\\ &=p^k + \chi(a_1\dots a_k)\lambda^k p^{\frac{k}{2}}\sum_{m\in\mathbb{F}_p^\times}e_p(-m)\chi(m)^k. \end{align*} The inner sum equals $-1$ or $\chi(-1)\lambda\sqrt{p}$ depending on whether $k$ is even or odd, therefore $$n=\begin{cases} p^{k-1}-\chi((-1)^{\frac{k}{2}}a_1\dots a_k)p^{\frac{k-2}{2}},&k\equiv 0\pmod{2};\\ p^{k-1}+\chi((-1)^{\frac{k-1}{2}}a_1\dots a_k)p^{\frac{k-1}{2}},&k\equiv 1\pmod{2}. \end{cases}$$

Yes, this is standard. More generally, we have the following

Theorem. Let $p$ be an odd prime, and let $a_1,\dots,a_k\in\mathbb{F}_p^\times$. Then the number of solutions of the quation $a_1x_1^2+\cdots +a_kx_k^2=1$ in $\mathbb{F}_p$ equals \begin{align*} p^{k-1}-\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 0\pmod{4};\\ p^{k-1}+\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 1\pmod{4};\\ p^{k-1}-\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 2\pmod{4};\\ p^{k-1}+\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 3\pmod{4}. \end{align*}

P.S. I am sure there is a reference for this, but I found it easier to copy from my notes.

Added. As Gerry Myerson pointed out, the special case $a_1=\dots=a_k=1$ is Proposition 8.6.1 in Ireland-Rosen: A classical introduction to modern number theory (2nd edition).

Yes, this is standard. More generally, we have the following

Theorem. Let $p$ be an odd prime, and let $a_1,\dots,a_k\in\mathbb{F}_p^\times$. Then the number of solutions of the quation $a_1x_1^2+\cdots +a_kx_k^2=1$ in $\mathbb{F}_p$ equals \begin{align*} p^{k-1}-\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 0\pmod{4};\\ p^{k-1}+\left(\frac{a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 1\pmod{4};\\ p^{k-1}-\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-2}{2}},&\qquad k\equiv 2\pmod{4};\\ p^{k-1}+\left(\frac{-a_1\dots a_k}{p}\right)p^{\frac{k-1}{2}},&\qquad k\equiv 3\pmod{4}. \end{align*}

P.S. I am sure there is a reference for this, but I found it easier to copy from my notes.

Added. As Gerry Myerson pointed out (in a comment he deleted), the special case $a_1=\dots=a_k=1$ is Proposition 8.6.1 in Ireland-Rosen: A classical introduction to modern number theory (2nd edition).