Are there integers m and n so that $\displaystyle m^{2} + (m+1)^{2} = n^{4} + (n + 1)^{4}$?
$\displaystyle m^2 + (m+1)^2 = n^4 + (n+1)^4 $
Let $\displaystyle m= a- \frac{1}{2} $ , $\displaystyle n = b-\frac{1}{2} $
Substitute them into the original equation :
$\displaystyle (a-\frac{1}{2})^2 + (a+\frac{1}{2})^2 = (b-\frac{1}{2})^4 + (b+\frac{1}{2})^4$
$\displaystyle 2a^2 + 2(\frac{1}{2})^2 = 2b^4 + 12b^2(\frac{1}{2})^2 + 2(\frac{1}{2})^4$
$\displaystyle a^2 + (\frac{1}{2})^2 = b^4 + \frac{3}{2} b^2 + \frac{1}{16} $
$\displaystyle a^2 + (\frac{1}{2})^2 = (b^2 + \frac{3}{4} )^2 + \frac{1}{16} - \frac{9}{16} $
Back substitution ,
$\displaystyle (m+\frac{1}{2})^2 + ( \frac{1}{4} + \frac{1}{2}) = [ (n+\frac{1}{2})^2 + \frac{3}{4} ]^2 $
$\displaystyle (2m+1)^2 + 3 = 4 ( n^2 + n + \frac{1}{4} + \frac{3}{4} )^2$
$\displaystyle (2m+1)^2 + 3 = 4 (n^2+n+1)^2 $
Put , owing to brevity $\displaystyle x = 2n^2 + 2n + 2 , ~y = 2m+1 $
we obtain $\displaystyle x^2 - y^2 = 3 $
$\displaystyle |x| = 2 , |y|=1 $
If
$\displaystyle n^2+n+1 = 1 $
$\displaystyle n= 0 $ or $\displaystyle n=-1$
$\displaystyle 2m+1 = 1$ or $\displaystyle 2m+1 = -1 $
$\displaystyle m = 0 $ or $\displaystyle n=-1$
Note that $\displaystyle n^2 + n + 1 = -1 $ doesn't have real solution so we conclude that $\displaystyle m,n $ are either $\displaystyle 0 $ or $\displaystyle -1 $ .