Let’s have the equation
$$(r_1\times p+p)^4-(4r_1^3+6r_1^2+4r_1+1)\times p^4=(r_1\times p)^4$$
And if $p=2$ and $r_1=n/2$ where $n$ a positive integer the above equation is written as follows:
$$(n+2)^4-[4(n/2)^3+6(n/2)^2+4(n/2)+1]\times2^4=n^4$$
In order to obtain formulas for Pythagorian quadruplets from this equation we set $n=[2\times(q^2+q)]$ and we have
$$[2(q^2+q+1)]^4-[64q^6+192q^5+288q^4+256q^3+160q^2+64q+16]=[2(n^2+n)]^4$$
But
$$64q^6+192q^5+288q^4+256q^3+160q^2+64q+16=[2(q+1)]^4+(8q^3+12q^2+8q)^2$$
and
$$64q^6+192q^5+288q^4+256q^3+160q^2+64q+16=(2q)^4+(8q^3+12q^2+8q+4)^2$$
After dividing by $16$ we obtain:
$$(q^2+q+1)^4=(q^2+q)^4+(q+1)^4+(2q^3+3q^2+2q)^2$$
and
$$(q^2+q+1)^4=(q^2+q)^4+q^4+(2q^3+3q^2+2q+1)^2$$
So we found two nontrivial solutions for the same greatest diagonal.
From my previous work on a different hyperelliptic equation I had found the following formula:
$$(q^2+q+1)^4=(q+1)^4+q^4+(q^4+2q^3+3q^2+2q)^2$$
If $q=2$ then we have $7^4=6^4+3^4+32^2$, $7^4=6^4+2^4+33^2$, $7^4=3^4+2^4+48^2$
So we have three nontrivial solutions.
Does anyone know how to obtain more than three nontrivial solutions in positive integers for the same greatest diagonal?
