So we have $2(m^2+m)+1=2(n^4+2n^3+3n^2+n)+1$ so $m(m+1)=n(n^3+2n^2+3n+1)$ and since $gcd(m,m+1)=gcd(n,n^3+2n^2+3n+1)=1$ then $m=n$ and $m+1=n^3+2n^2+3n+1$? then we get $m=(0,-1)$ & $n=(0,-1)$

Imgoodinmaths_dude wrote: So we have $2(m^2+m)+1=2(n^4+2n^3+3n^2+n)+1$ so $m(m+1)=n(n^3+2n^2+3n+1)$ and since $gcd(m,m+1)=gcd(n,n^3+2n^2+3n+1)=1$ then $m=n$ and $m+1=n^3+2n^2+3n+1$? then we get $m=(0,-1)$ & $n=(0,-1)$ Imgoodinmaths_dude, I think it should be $2(m^2+m)+1= 2(n^4+2n^3+3n^2+2n)+1,$ which is equivalent to $(2n^2+2n-2m+1)(2n^2+2n+2m+3)=3$.