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.