Hi
Total number of integer ordered pair
\[ (a,b) \] in
\[ a^2+a=b^4+b^3+b^2+b \] (*)
You have:
\( a^2+a=(a+1/2)^2-1/4 \)
\( b^4+b^3+b^2+b=(b^2+b/2+1/2)^2-(b^2/4-b/2)-1/4 \)
So, if we multiply (*) by \( 4 \), we obtain:
\( (2a+1)^2=(2b^2+b+1)^2-(b^2-2b) \)
\( (2b^2+b+1)^2-(2a+1)^2=b^2-2b \) (**)
If \( b^2-2b> 0 \) (that is \( b> 2 \) or \( b< 0 \)), then
\( (2b^2+b+1)^2-(2a+1)^2>0 \)
Let us see the difference of \( (2b^2+b+1)^2 \) with the nearest square \( (2b^2+b)^2 \):
\( (2b^2+b+1)^2-(2b^2+b)^2=4b^2+2b+1=b^2-2b+(3b^2+4b+1)>b^2-2b \) if \( 3b^2+4b+1>0 \).
But for integer values \( 3b^2+4b+1>0 \) if \( b\neq 0,-1 \) and in these cases equiality (**) is impossible.
So we only have to check the cases \( b=-1,0,1,2 \).
- For \( b=0,-1 \) the equation (*) is \( a(a+1)=0 \), that is, \( a=0 \) or \( a=-1 \). We obtain the pairs \( (0,-1),(-1,-1),(0,0),(-1,0) \).
- For \( b=1 \) the equation (*) is \( a(a+1)=4 \). There are not integer solutions.
- For \( b=2 \) the equation (*) is \( a(a+1)=30 \), that is, \( a=5 \) or \( a=-6 \). We obtain the pairs \( (5,2),(-6,2) \).
Best regards.
