This article shows that the only solutions in integers to the equation
are the obvious trivial solutions together with . This result was known to Fermat.
First, note that the equation is , so we have immediately that . So, noting the solutions for , assume in what follows that .
Let . Then , so that . But so that and is either or . If , so that and are coprime, then and must both be squares. But is not a square for . Thus the of and is , so each must be twice a square, say
so that form a primitive Pythagorean triple (note that since ).
Recall that if is a primitive Pythagorean triple, then precisely one of and is even, and we can choose coprime integers such that or depending on the parity of .
Suppose first that is odd. Then
Then . Now, note that must be a square, say , since and is a square. Then
So suppose that is even. Then
From the second of these formulas, we see that must be even (consider both sides ), say . Now,
Since and have opposite parity, are even, so that
Thus one of is a fourth power and the other is twice a fourth power, say and .
so that .
If , then
and again we have a square being the sum of two fourth powers. So this case is impossible.
If , write , then . It follows (see here (http://planetmath.org/X4Y4z2HasNoSolutionsInPositiveIntegers)) that either or . If , we get the impossibility , while if , then and so . Then , so that . Thus , so and, finally, , so that .
Thus, the only nontrivial solution to the equation given is
|Date of creation||2013-03-22 17:05:09|
|Last modified on||2013-03-22 17:05:09|
|Last modified by||rm50 (10146)|