How would you solve the diophantine $x^4+y^4=2z^2$

Question

I would like to know any way of solving the diophantine equation $x^4+y^4=2z^2$. Or ideas that seem worth trying out.

By solving I mean fining all solutions and proving there are no more.

Keith Conrad showed how to reduce this equation to a different one which was solved by Fermat in his notes about Fermat descent, other than I have no ideas how to solve it. I tried to do descent on it directly but that seems completely impossible so I am interested in other techniques. Thanks very much.

page 451 of Number Theory: Analytic and modern tools by Henri Cohen shows how to reduce it to an elliptic curve. — quanta, Mar 12 '11 at 22:05

Derek Jennings · Accepted Answer · 2011-03-12T20:38:13.613

3

A quick spot of googling found Two fourths and a square.

Look at pattern six: the unique coprime solution is $1^4+1^4=2 \cdot 1^2 .$

edited Mar 12 '11 at 20:38

answered Mar 12 '11 at 19:57

Derek Jennings

13,661

Strictly that is the unique coprime solution. So the complete solution is $(x,y,z) = (n, \pm n, \pm n^2)$ for any integer $n$ – Henry Mar 12 '11 at 20:16
@Henry: Yes, I meant coprime solution, though I should have stated it explicitly. – Derek Jennings Mar 12 '11 at 20:38

How would you solve the diophantine $x^4+y^4=2z^2$

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