There is an elementary argument regarding the last problem. Denote by $N(p)$ the number of pairs of $(a,b)$ such that $a,b,a+b$ are all quadratic residues mod $p$. Hence we have $$N(p)=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab(a+b),p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$

$$=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$

$$-\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1,p|a+b}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right).$$

Clearly, the second term is just \begin{align*}&-\frac{1}{8}\sum_{\substack{a\bmod p\\(a,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{-a}{p}\right)\right)=\frac{1}{8}-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right).\end{align*} And for the first term, we are required to investigate the quantity \begin{align*}L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ab(a+b)}{p}\right).\end{align*} In fact we have

$$L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ba^2+b^2a}{p}\right) =\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{b(a+\overline{2}b)^2-\overline{4}b^3}{p}\right)$$

$$=\mathop{\sum\sum}_{a,b\bmod p}\left(\frac{ba^2-\overline{4}b^3}{p}\right)$$

$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-\overline{4}b^2}{p}\right)$$

$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-1}{p}\right)=0.$$

The other terms could be computed in a similar way. Hence we can deduce that \begin{align*}N(p)=\frac{1}{8}(p-1)^2-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right)+\frac{1}{8}.\end{align*}

There is an elementary argument regarding the last problem. Denote by $N(p)$ the number of pairs of $(a,b)$ such that $a,b,a+b$ are all quadratic residues mod $p$. Hence we have $$N(p)=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab(a+b),p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$

$$=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$

$$-\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1,p|a+b}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right).$$

Clearly, the second term is just \begin{align*}&-\frac{1}{8}\sum_{\substack{a\bmod p\\(a,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{-a}{p}\right)\right)=\frac{1}{8}-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right).\end{align*} And for the first term, we are required to investigate the quantity \begin{align*}L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ab(a+b)}{p}\right).\end{align*} In fact we have

$$L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ba^2+b^2a}{p}\right) =\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{b(a+\overline{2}b)^2-\overline{4}b^3}{p}\right)$$

$$=\mathop{\sum\sum}_{a,b\bmod p}\left(\frac{ba^2-\overline{4}b^3}{p}\right)$$

$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-\overline{4}b^2}{p}\right)$$

$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-1}{p}\right)=0.$$

The other terms could be computed in a similar way. Hence we can deduce that \begin{align*}N(p)=\frac{1}{8}(p-1)^2-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right)+\frac{1}{8}.\end{align*}

There is an elementary argument regarding the last problem. Denote by $N(p)$ the number of pairs of $(a,b)$ such that $a,b,a+b$ are all quadratic residues mod $p$. Hence we have \begin{align*}N(p)&=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab(a+b),p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)\\ &=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)\\ &\ \ \ \ \ \ -\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1,p|a+b}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right).\end{align*}

Clearly, the second term is just \begin{align*}&-\frac{1}{8}\sum_{\substack{a\bmod p\\(a,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{-a}{p}\right)\right)=\frac{1}{8}-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right).\end{align*} And for the first term, we are required to investigate the quantity \begin{align*}L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ab(a+b)}{p}\right).\end{align*} In fact we have \begin{align*}L:&=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ba^2+b^2a}{p}\right) =\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{b(a+\overline{2}b)^2-\overline{4}b^3}{p}\right)\\ &=\mathop{\sum\sum}_{a,b\bmod p}\left(\frac{ba^2-\overline{4}b^3}{p}\right)\\ &=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-\overline{4}b^2}{p}\right)\\ &=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-1}{p}\right)\\ &=0.\end{align*}

The other terms could be computed in a similar way. Hence we can deduce that \begin{align*}N(p)=\frac{1}{8}(p-1)^2-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right)+\frac{1}{8}.\end{align*}

There is an elementary argument regarding the last problem. Denote by $N(p)$ the number of pairs of $(a,b)$ such that $a,b,a+b$ are all quadratic residues mod $p$. Hence we have $$N(p)=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab(a+b),p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$

$$=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$

$$-\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1,p|a+b}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right).$$

Clearly, the second term is just \begin{align*}&-\frac{1}{8}\sum_{\substack{a\bmod p\\(a,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{-a}{p}\right)\right)=\frac{1}{8}-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right).\end{align*} And for the first term, we are required to investigate the quantity \begin{align*}L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ab(a+b)}{p}\right).\end{align*} In fact we have

$$L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ba^2+b^2a}{p}\right) =\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{b(a+\overline{2}b)^2-\overline{4}b^3}{p}\right)$$

$$=\mathop{\sum\sum}_{a,b\bmod p}\left(\frac{ba^2-\overline{4}b^3}{p}\right)$$

$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-\overline{4}b^2}{p}\right)$$

$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-1}{p}\right)=0.$$

The other terms could be computed in a similar way. Hence we can deduce that \begin{align*}N(p)=\frac{1}{8}(p-1)^2-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right)+\frac{1}{8}.\end{align*}