## Diophantine Equation With No Solutions

Theorem (Diophantine Equation With No Solutions)
The equation
$x^4+y^4=z^2$
has no integer solutions.
Proof
Suppose
$x_1, \; y_1, \; z_1$
is a solution. Suppose
$gcd(x, \; y)=d \gt 1$
then
$x^4+y^4=z^2 \rightarrow (dx')^4+(yd')^4=z^2 \rightarrow d^2 | z$
and
$gcd(x', \; y' )=1$
. We can assume therefore that
$gcd(x', \; y' )=1$
. By writing the equation as
$(x_1^2)^2+(y_1^2)^2=z_1^2$
. (Pythagoras) we can take
$x_1^2, \; y_1^2, \; z_1$
as Pythagorean Triples,
$x_1=2mn,, y^2_1=m^2-n^2$
, where
$gcd(m, \; n)=1, \; m \gt n /gt 0$
, and
$m, \; n$
have opposite polarity.
$m, \; n$
must be odd and even respectively since if the other way round then
$(2k)^2-(2j+1)^2=4k^2-4j^2-4j-1 \equiv 3 \; (mod \; 4)$
but any square number is equal to 0 or 1 modulo 4.
Putting
$m=2r$
the equation for
$x_1$
becomes
$$$\frac{x_1}{2}$$^2=mr$
where
$m$
and
$r= \frac{n}{2}$
are relatively prime, hence
$m, \; r$
are both square numbers. Put
$m=s^2, \; r=t^2$
>
From the equation
$y_2^2=m^2-n^2$

$(y_1, \; m, \; n)$
is a Pythagorean triple so
$n=2uv, \; y_1=u^2-v^2, \; m=u^2+v^2$
where
$u, \; v$
are relatively prime integers off opposite polarity with
$u \gt v$
.
Now
$n=2r=2t^2$
so the first of these equations becomes
$uv=t^2$
. It follows that
$u, \; v$
are each square numbers say
$u=x_2^2, v=y_2^2$
and substituting into
$m=u^2+v^2$
gives
$x^2_2+y^2_2=m=s^2 \; ({}=z^2_2)$
say. This solution satisfies
$0 \lt z_2=s \le m \lt m^2+n^2=z_1$
.
Hence a smaller solution exists, which is impossible and the proof is complete.