1. ## diophintine equation perhaps?

Are there integers m and n so that $m^{2} + (m+1)^{2} = n^{4} + (n + 1)^{4}$?

2. m=n=0

3. Originally Posted by FancyMouse
m=n=0

And $m=-1\,,\,n=0\,,\,\,m=0\,,\,n=-1$ ...

Tonio

4. $m^2 + (m+1)^2 = n^4 + (n+1)^4$

Let $m= a- \frac{1}{2}$ , $n = b-\frac{1}{2}$

Substitute them into the original equation :

$(a-\frac{1}{2})^2 + (a+\frac{1}{2})^2 = (b-\frac{1}{2})^4 + (b+\frac{1}{2})^4$

$2a^2 + 2(\frac{1}{2})^2 = 2b^4 + 12b^2(\frac{1}{2})^2 + 2(\frac{1}{2})^4$

$a^2 + (\frac{1}{2})^2 = b^4 + \frac{3}{2} b^2 + \frac{1}{16}$

$a^2 + (\frac{1}{2})^2 = (b^2 + \frac{3}{4} )^2 + \frac{1}{16} - \frac{9}{16}$

Back substitution ,

$(m+\frac{1}{2})^2 + ( \frac{1}{4} + \frac{1}{2}) = [ (n+\frac{1}{2})^2 + \frac{3}{4} ]^2$

$(2m+1)^2 + 3 = 4 ( n^2 + n + \frac{1}{4} + \frac{3}{4} )^2$

$(2m+1)^2 + 3 = 4 (n^2+n+1)^2$

Put , owing to brevity $x = 2n^2 + 2n + 2 , ~y = 2m+1$

we obtain $x^2 - y^2 = 3$

$|x| = 2 , |y|=1$

If

$n^2+n+1 = 1$
$n= 0$ or $n=-1$

$2m+1 = 1$ or $2m+1 = -1$
$m = 0$ or $n=-1$

Note that $n^2 + n + 1 = -1$ doesn't have real solution so we conclude that $m,n$ are either $0$ or $-1$ .