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.
