Show that, there are no integers $(m, n)$ such that, $m^2 + (m + 1)^2 = n^4 + (n + 1)^4$.
Okay so i see that $m>n$ as smaller powers of $m$ are being to compared to larger powers of $n$. So my first thought was turning the equation into something of the form $negative=positive$ which is a contradiction hence no solutions. Intuitively i feel $m>n^2$ for the same reason mentioned previously. But idk how we can establish that claim.
Show that $x^4 + y^4 - z^4 = 1993$ has no solutions in integers.
Well this one my first thought was congruences and parity (odd/even) as all 4th powers are $\equiv 1 $ modulos $2,3,4,8$ (that is only if they are not divisble by those numbers, point is its either $0$ or $1$.). But well that doesn't seem like the way to start the question though it certainly seems to be useful midway. For beginning the question i thought of maybe rewriting the thing as $x^4 - z^4 = 1993-y^4$ and factor but i dont think that leads anywhere.
Could someone provide beautiful/elegant solutions which helps me learn and understand how to approach such questions?
Rewrite the first equation in the form: $$m^2+m+1=(n^2+n+1)^2.$$ Can you end it now?
I got that it's possible for $m^2+m+1=1$ only.