There is no solution...
to the Diophantine equation A4+B4=C2.
Note that this is the first stage of proving Fermat's Last theorem for n=4.
Here, A2,B2,C are Primitive Pythagorean Triple
(assuming without the loss of generality, B2 is even), so
A2=a2-b2
B2=2ab
C=a2+b2
Rearrange the first equation to get A2+b2=a2,
so A,b,a is a Pythagorean Triple with b even. Thus
A=c2-d2
b=2cd
a=c2+d2
Substitute the expressions for b and a back to B2=2ab, so we have
B2=4cd(c2+d2)
As c, d, c2+d2 pairwise coprime, each of them must be a
perfect square. Therefore,
c=e2, d=f2 and c2+d2=g2
Combine these expression to give
e4+f4=g2
with g £ g2 = a £
a2 < C
Therefore, assuming C is minimally chosen on the outset, we found a smaller
value for C. Thus, an infinite descent is produced - no solution as
required.
Finally, can you see how to use this information to prove no solution for
Fermat's equation with n=4?
Related Pages in this website
Infinite Descent Proofs
Pythagorean Triples
Triangles
Formulas for Primitive Pythagorean
Triples and their deriviation -- a way to generate all the triples such
that a^2 + b^2 = c^2
Prove that the area
of a right triangle with integer sides is not a perfect square. The
proof is here (with some help from someone from the nrich website)
Puzzle
question -- if m is the product of n distinct primes, how many different
right triangles with integer sides have a leg of length m?
Arithmetic Sequence of Perfect
Squares, page 3 -- If a^2, b^2, c^2 are in arithmetic sequence, why is
their constant difference a multiple of 24? Look at the second answer to
this question for the relationship between Pythagorean Triples and this
arithmetic sequence of squares.
Theorems Involving Perfect Squares
-- answers questions such as "why is the square root of x irrational
unless x is a perfect square?" and other fundamental questions about
perfect squares.
Fermat's Therems.
