|
|
|
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 coprime integers a,b exist such that
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 coprime integers c,d exist such that
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?
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.
The webmaster and author of this Math Help site is Graeme McRae.