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This came up with the Euler brick.

Let $(p,q,r)$ be a Randall triple, i.e. $$(p^2-1)(q^2-1)(r^2-1)=8pqr\ \qquad\text{[eq.1]}.$$ There are tons of maps that map a triple to another: obvious $$p\mapsto -p,\;q\mapsto 1/q,\;r\mapsto r,$$ still easy $$p\mapsto (p+1)/(p-1),\;q\mapsto (q+1)/(q-1),\;r\mapsto (r+1)/(r-1))$$ and outrageous $$p\mapsto (\dots+p^2q^2r^2)/(\dots+p^2q^2r^2),$$ likewise $q,r$ - I found three such maps).

Thus let $F[p,q,r]$ be a broken rational function of $p,q,r$ where the maximum exponent of each variable in numerator and denominator is $2$ (so overall $6/6$), and $p'=F[p,q,r],q'=F[q,r,p],r'=F[r,p,q]$. This makes up for a whopping $54$ free coefficients, so it is hopeless to check if $p',q',r'$ fulfils Eq.1 by plugging it in, do a resultant with Eq.1 to eliminate $r$ and set the coefficients (estimated: up to $p^{100}$) of $p,q$ to zero.

My maps are not only birational, but even involutions. Are there indicators for "F is birational" or "F is even an involution" that help eliminating a bucketload of coefficients first?

The following might help to massacre a few coefficients beforehand: $T=(0,1,s)$ is a valid triple for any $s$, so $F$ must map $T$ either to the other $23$ triples that can be obtained by using the above trivial maps on $T$, or $F$ must give $(p',q',r')=(0/0,0/0,0/0)$ (or another indeterminate) which evaluates to a meaningful limit which is a parametric solution (e.g. the Saunderson one).

This came up with the Euler brick.

Let $T=(p,q,r)$ be a Randall triple, i.e. $$(p^2-1)(q^2-1)(r^2-1)=8pqr\ \qquad\text{[eq.1]}.$$ There are tons of maps that map a triple $T$ to another $T'=(p',q',r')$: obvious $$p\mapsto -p,\;q\mapsto 1/q,\;r\mapsto r,$$ still easy $$p\mapsto (p+1)/(p-1),\;q\mapsto (q+1)/(q-1),\;r\mapsto (r+1)/(r-1))$$ and outrageous $$p\mapsto (\dots+p^2q^2r^2)/(\dots+p^2q^2r^2),$$ ($q',r'$ shall be cyclic - I found three such maps).

EDIT: For your convenience, a concrete example is $$p\mapsto -(((-1+q)*(1+p+q+p*q+2*p*r-2*p*q*r-r^2+p*r^2-q*r^2+p*q*r^2))/((1+q)*(1+p-q-p*q+2*r+2*q*r+r^2-p*r^2-q*r^2+p*q*r^2)))$$ which also takes $0,-1,s$ to the Saunderson solution, and also is an identity (so to say! - it's a combination of the easy maps above) on $-p,-q,-r$ (which is no allowed map for Eq.1!). Note that the degrees are not maximal here, and don't have to be, thus the $\dots p^2q^2r^2$ notation above.

Thus let $F[p,q,r]$ be a broken rational function of $p,q,r$ (like in the last map example) where the maximum exponent of each variable in numerator and denominator is $2$ (so overall $6/6$), and $p'=F[p,q,r],q'=F[q,r,p],r'=F[r,p,q]$. This makes up for a whopping $54$ free coefficients, so it is hopeless to check if $p',q',r'$ fulfils Eq.1 by plugging it in, do a resultant with Eq.1 to eliminate $r$ and set the coefficients (estimated: up to $p^{100}$) of $p,q$ to zero.

My maps are not only birational, but even involutions. Are there indicators for "F is birational" or "F is even an involution" that help eliminating a bucketload of coefficients first?

The following might help to massacre a few coefficients beforehand: $T=(0,1,s)$ is a valid triple for any $s$, so $F$ must map $T$ either to the other $23$ triples that can be obtained by using the above trivial maps on $T$, or $F$ must give $(p',q',r')=(0/0,0/0,0/0)$ (or another indeterminate) which evaluates to a meaningful limit which is a parametric solution (e.g. the Saunderson one).

This came up with the Euler brick.

Let $(p,q,r)$ be a Randall triple, i.e. $$(p^2-1)(q^2-1)(r^2-1)=8pqr\ \qquad\text{[eq.1]}.$$ There are tons of maps that map a triple to another: obvious $$p\rightarrow -p,\;q\rightarrow 1/q,\;r\rightarrow r,$$ still easy $$p\rightarrow (p+1)/(p-1),\;q\rightarrow (q+1)/(q-1),\;r\rightarrow (r+1)/(r-1))$$ and outrageous $$p\rightarrow (\dots+p^2q^2r^2)/(\dots+p^2q^2r^2),$$ likewise $q,r$ - I found three such maps).

Thus let $F[p,q,r]$ be a broken rational function of $p,q,r$ where the maximum exponent of each variable in numerator and denominator is $2$ (so overall $6/6$), and $p'=F[p,q,r],q'=F[q,r,p],r'=F[r,p,q]$. This makes up for a whopping $54$ free coefficients, so it is hopeless to check if $p',q',r'$ fulfils Eq.1 by plugging it in, do a resultant with Eq.1 to eliminate $r$ and set the coefficients (estimated: up to $p^{100}$) of $p,q$ to zero.

My maps are not only birational, but even involutions. Are there indicators for "F is birational" or "F is even an involution" that help eliminating a bucketload of coefficients first?

The following might help to massacre a few coefficients beforehand: $T=(0,1,s)$ is a valid triple for any $s$, so $F$ must map $T$ either to the other $23$ triples that can be obtained by using the above trivial maps on $T$, or $F$ must give $(p',q',r')=(0/0,0/0,0/0)$ (or another indeterminate) which evaluates to a meaningful limit which is a parametric solution (e.g. the Saunderson one).

This came up with the Euler brick.

Let $(p,q,r)$ be a Randall triple, i.e. $$(p^2-1)(q^2-1)(r^2-1)=8pqr\ \qquad\text{[eq.1]}.$$ There are tons of maps that map a triple to another: obvious $$p\mapsto -p,\;q\mapsto 1/q,\;r\mapsto r,$$ still easy $$p\mapsto (p+1)/(p-1),\;q\mapsto (q+1)/(q-1),\;r\mapsto (r+1)/(r-1))$$ and outrageous $$p\mapsto (\dots+p^2q^2r^2)/(\dots+p^2q^2r^2),$$ likewise $q,r$ - I found three such maps).

Thus let $F[p,q,r]$ be a broken rational function of $p,q,r$ where the maximum exponent of each variable in numerator and denominator is $2$ (so overall $6/6$), and $p'=F[p,q,r],q'=F[q,r,p],r'=F[r,p,q]$. This makes up for a whopping $54$ free coefficients, so it is hopeless to check if $p',q',r'$ fulfils Eq.1 by plugging it in, do a resultant with Eq.1 to eliminate $r$ and set the coefficients (estimated: up to $p^{100}$) of $p,q$ to zero.

My maps are not only birational, but even involutions. Are there indicators for "F is birational" or "F is even an involution" that help eliminating a bucketload of coefficients first?

The following might help to massacre a few coefficients beforehand: $T=(0,1,s)$ is a valid triple for any $s$, so $F$ must map $T$ either to the other $23$ triples that can be obtained by using the above trivial maps on $T$, or $F$ must give $(p',q',r')=(0/0,0/0,0/0)$ (or another indeterminate) which evaluates to a meaningful limit which is a parametric solution (e.g. the Saunderson one).

This came up with the Euler brick.

Let $(p,q,r)$ be a Randall triple, i.e. $(p^2-1)(q^2-1)(r^2-1)=8pqr\ [eq.1]$. There are tons of maps that map a triple to another: obvious ($p\rightarrow -p,q\rightarrow 1/q,r\rightarrow r)$, still easy ($p\rightarrow (p+1)/(p-1),q\rightarrow (q+1)/(q-1),r\rightarrow (r+1)/(r-1))$ and outrageous ($p\rightarrow (\dots+p^2q^2r^2)/(\dots+p^2q^2r^2)$, likewise $q,r$ - I found three such maps).

Thus let $F[p,q,r]$ be a broken rational function of $p,q,r$ where the maximum exponent of each variable in numerator and denominator is $2$ (so overall $6/6$), and $p'=F[p,q,r],q'=F[q,r,p],r'=F[r,p,q]$. This makes up for a whopping $54$ free coefficients, so it is hopeless to check if $p',q',r'$ fulfils Eq.1 by plugging it in, do a resultant with Eq.1 to eliminate $r$ and set the coefficients (estimated: up to $p^{100}$) of $p,q$ to zero.

My maps are not only birational, but even involutions. Are there indicators for "F is birational" or "F is even an involution" that help eliminating a bucketload of coefficients first?

The following might help to massacre a few coefficients beforehand: $T=(0,1,s)$ is a valid triple for any $s$, so $F$ must map $T$ either to the other $23$ triples that can be obtained by using the above trivial maps on $T$, or $F$ must give $(p',q',r')=(0/0,0/0,0/0)$ (or another indeterminate) which evaluates to a meaningful limit which is a parametric solution (e.g. the Saunderson one).

This came up with the Euler brick.

Let $(p,q,r)$ be a Randall triple, i.e. $$(p^2-1)(q^2-1)(r^2-1)=8pqr\ \qquad\text{[eq.1]}.$$ There are tons of maps that map a triple to another: obvious $$p\rightarrow -p,\;q\rightarrow 1/q,\;r\rightarrow r,$$ still easy $$p\rightarrow (p+1)/(p-1),\;q\rightarrow (q+1)/(q-1),\;r\rightarrow (r+1)/(r-1))$$ and outrageous $$p\rightarrow (\dots+p^2q^2r^2)/(\dots+p^2q^2r^2),$$ likewise $q,r$ - I found three such maps).

Thus let $F[p,q,r]$ be a broken rational function of $p,q,r$ where the maximum exponent of each variable in numerator and denominator is $2$ (so overall $6/6$), and $p'=F[p,q,r],q'=F[q,r,p],r'=F[r,p,q]$. This makes up for a whopping $54$ free coefficients, so it is hopeless to check if $p',q',r'$ fulfils Eq.1 by plugging it in, do a resultant with Eq.1 to eliminate $r$ and set the coefficients (estimated: up to $p^{100}$) of $p,q$ to zero.

My maps are not only birational, but even involutions. Are there indicators for "F is birational" or "F is even an involution" that help eliminating a bucketload of coefficients first?

The following might help to massacre a few coefficients beforehand: $T=(0,1,s)$ is a valid triple for any $s$, so $F$ must map $T$ either to the other $23$ triples that can be obtained by using the above trivial maps on $T$, or $F$ must give $(p',q',r')=(0/0,0/0,0/0)$ (or another indeterminate) which evaluates to a meaningful limit which is a parametric solution (e.g. the Saunderson one).