I may be found in . Some of my favourite original identities: $$(acdot x+bcdot y)^2+(acdot overline{a_1+p}+bcdotoverline{b_1+q})^2+(acdot overline{a_2+p}+bcdotoverline{b_2+q})^2$$ $$+(acdotoverline{a_1+a_2+x}+bcdotoverline{b_1+b_2+y})^2$$ $$=$$ $$(acdot p+bcdot q)^2+(acdotoverline{a_1+x}+bcdotoverline{b_1+y})^2 +(acdotoverline{a_2+x}+bcdotoverline{b_2+y})^2$$ $$+(acdotoverline{a_1+a_2+p}+bcdotoverline{b_1+b_2+q})^2$$ $$bowtie$$ If $ab+bc+ac=0$, then $$(2a+b)^2+(2b+c)^2+(2a+c)^2=(2a-b)^2+(2b-c)^2+(2a-c)^2$$ $$(-a+b+c)^2+(a-b+c)^2+(a+b-c)^2=3(a+b+c)^2$$ $$frac{({-3a}+b+c)^5+(a-3b+c)^5+(a+b-3c)^5+(a+b+c)^5}{({-3a}+b+c)^3+(a-3b+c)^3+(a+b-3c)^3+(a+b+c)^3}=10(a+b+c)^2$$ $$bowtie$$ For all $qnotin [-1,1]$, it follows for the $n^{text{th}}$ $p_n$ that $$sum_{n=1}^{infty}bigg(frac 1{q^{p_n}-1}-sum_{k=1}^{n-1}frac 1{q^{p_np_{n-k}}-1}bigg)=frac 1{q(q-1)}$$ $$bowtie$$ $$sqrt[3]{frac 52bigg(frac{sqrt[5]2+4}{sqrt[5]4+2}-1bigg)}-sqrt[3]{frac 7{sqrt[5]8}-frac{2-sqrt[5]2}{2+sqrt[5]8}-4}=sqrt[5]2-sqrt[5]{frac 14}$$ $$frac{7-sqrt 5+sqrt{6big(5-sqrt 5big)}}{7+sqrt 5-sqrt{6big(5+sqrt 5big)}}=frac 14bigg(9-sqrt 5+sqrt{6cdotoverline{5+sqrt 5}}bigg)$$ $$sqrt[4]{3bigg(1-frac 1{sqrt[3]4}bigg)}Bigg(sqrt{frac 3{sqrt[3]2}-1}+sqrt{frac 13cdotoverline{1+frac 1{sqrt[3]2}}}Bigg)=2$$ $$bowtie$$ Denote by $Phi_n(x)$ the , then for all odd $pn$ given $p$ is prime, it follows that $$cfrac{Phi_{pn}(x)+Phi_{pn}(-x)}{Phi_{pn}(x)-Phi_{pn}(-x)}=cfrac{Phi_{pn}big(frac 1xbig)+Phi_{pn}big({-frac 1x}big)}{Phi_{pn}big(frac 1xbig)-Phi_{pn}big({-frac 1x}big)}$$ $$bowtie$$ If $tneq 0$ is a root of $(b-at)u^2+2(c-bt)u+3(d-ct)=0$, then $dfrac 1t$ is a root of $$frac 14big(a+bv+cv^2+dv^3big)=a+(b-at)v+(c-bt)v^2+(d-ct)v^3$$ $$bowtie$$ For $aneq 0$, let $r=sqrt[3]{dfrac{b^2-ac}{a^2}}$ and $t=dfrac ba+1$. If $$d=-frac 1a(c-bt)^2+ct$$ then $$ax^3+3bx^2+3cx+d=0$$ with $x=r^2+r-dfrac ba$. $$bowtie$$ If $x=dfrac{a_1}{a_2}+dfrac{b_1}{b_2}$ then, for any $a$ and $b$, it follows that $$(a_2x^2-a_1x+a)(b_2x^2-b_1x+b)=(a_2b+a_1b_1+ab_2)x^2-(ab_1+a_1b)x+ab$$ $$bowtie$$ If $varphi$ is the and $psi$ is its , then $$frac{pi}5sqrt{frac{2-psi}{2+psi}}$$ $$=bigg(1-frac{2-varphi}{2+varphi}bigg)bigg(1-cfrac{2-sqrt{2+varphi}}{2+sqrt{2+varphi}}bigg)Bigg(1-cfrac{2-sqrt{2+sqrt{2+varphi}}}{2+sqrt{2+sqrt{2+varphi}}}Bigg)cdots$$ $$bowtie$$ $$2arctanfrac 1{sqrt[3]2}-arctanfrac 5{3+4sqrt[3]2}=frac{pi}4$$ ©