On $119^2+120^2=13^4\,$ and $p=239$ We are familiar with, $$\frac\pi4=4\arctan\tfrac15-\arctan\tfrac1{239}$$Let $p=a+b=239$ and $(a,b,c,d)=(120,119,13,2).\,$ Some years back, I observed this rather long list of Diophantine relations and Pell equations that they satisfied,
$$120^2-119^2 = 239$$
$$120^2 + 119^2 = 13^4$$
$$239^2+(13^4-1)^2=13^8$$
$$239^2 - 2\cdot13^4 = -1$$
$$13^4 - \frac{119\times120}2\,2^2 = 1$$
$$1^3+3^3+5^3+\dots+(2\cdot13^2-1)^3 = (13^2\cdot239)^2$$
$$1^3+2^3+3^3 +\dots+(239^2)^3 = (13^2\cdot239)^4$$
$$120^4 + 13^4+ (2\cdot13)^4 = 119^4 + (4\cdot13)^4$$
and this new post just adds to the list,
$$\frac{u(u+1)(u+2)(u+3)}{8} = \frac{v(v+1)(v+2)}{6} = \frac{w(w+1)}{2} = \frac{119\times120}2=7140$$ 
with $u=2\times\color{blue}7,\;v=2\times\color{blue}{17},\; w= \color{blue}{7\times17} =119$ such that $7140$ is the largest number that is both triangular and tetrahedral. (And is $3\times$ the pentatope number $2380$). 

Q: Any other non-trivial Diophantine relations involving $2$nd and higher powers that $a,b,c,d,p$ satisfies?


Another relation is:
$$(2\cdot2)^4-2^4-239=1$$
This implies:
$$239\cdot13^4 + 13^4 + (2\cdot13)^4 = (2\cdot2\cdot13)^4$$
which, using the first two relations on your list, gives:
$$(120^2-119^2)(120^2+119^2)+13^4+(2\cdot13)^4=(2\cdot2\cdot13)^4$$
which is equivalent to the last relation on your list.
