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Since (y 2, z, x 2) form a primitive Pythagorean triple, they can be written z = 2de y 2 = d 2 − e 2 x 2 = d 2 + e 2. where d and e are coprime and d > e > 0. Thus, x 2 y 2 = d 4 − e 4. which produces another solution (d, e, xy) that is smaller (0 < d < x). As before, there must be a lower bound on the size of solutions, while this argument ...
Equivalent statement 2: x n + y n = z n, where integer n ≥ 3, has no non-trivial solutions x, y, z ∈ Q. This is because the exponents of x , y , and z are equal (to n ), so if there is a solution in Q , then it can be multiplied through by an appropriate common denominator to get a solution in Z , and hence in N .
Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes. [1] [2]Mathematical induction is a method for proving that a statement () is true for every natural number, that is, that the infinitely many cases (), (), (), (), … all hold.
The Fermat numbers satisfy the following recurrence relations: = + = + for n ≥ 1, = + = for n ≥ 2.Each of these relations can be proved by mathematical induction.From the second equation, we can deduce Goldbach's theorem (named after Christian Goldbach): no two Fermat numbers share a common integer factor greater than 1.
On the other hand, the primes 3, 7, 11, 19, 23 and 31 are all congruent to 3 modulo 4, and none of them can be expressed as the sum of two squares. This is the easier part of the theorem, and follows immediately from the observation that all squares are congruent to 0 (if number squared is even) or 1 (if number squared is odd) modulo 4.
For example, we can prove by induction that all positive integers of the form 2n − 1 are odd. Let P(n) represent "2n − 1 is odd": (i) For n = 1, 2n − 1 = 2(1) − 1 = 1, and 1 is odd, since it leaves a remainder of 1 when divided by 2. Thus P(1) is true.
Euler was aware of the equality 59 4 + 158 4 = 133 4 + 134 4 involving sums of four fourth powers; this, however, is not a counterexample because no term is isolated on one side of the equation. He also provided a complete solution to the four cubes problem as in Plato's number 3 3 + 4 3 + 5 3 = 6 3 or the taxicab number 1729.
The argument is proof by induction. First, we establish a base case for one horse ( n = 1 {\displaystyle n=1} ). We then prove that if n {\displaystyle n} horses have the same color, then n + 1 {\displaystyle n+1} horses must also have the same color.