In other words, to prove 2^{99} + 1 is composite we constructed a factorization. Not surprisingly, we call such a proof **constructive**.

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**Theorem.** There is a two parameter family of Pythagorean triples.

**Proof.** (We construct the solution.) Let a = u^{2} - v^{2} and b = 2 u v where u and v are positive integers with u > v. Then a^{2} + b^{2} = (u^{2}-v^{2})^{2} + (2uv)^{2} = u^{4} - 2 u^{2}v^{2} + v^{4} + 4 u^{2}v^{2} = u^{4} + 2u^{2}v^{2} + v^{4} = (u^{2} + v^{2})^{2}. Thus, (u^{2} - v^{2}, 2 u v, u^{2} + v^{2}), for u > v, is two parameter family of Pythagorean triples.

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**Theorem.** There is a rational number that lies strictly between the square root of 10^{100} and the square root of 10^{100}+1.

**Proof.** The square root of 10^{ 100} is 10^{ 50}. After a little bit of trial and error, we let x = 10^{ 50} + 10^{ -51}, which is clearly a rational number bigger than the squre root of 10^{ 100}. To prove that x is less than the square root of 10^{100}+1, we compute

x^{2} = (10^{ 50} + 10^{ -51})^{2} = 10^{ 100} + (2) 10^{ -1} + 10^{ -102}

which is clearly less than 10^{100}+1.

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**Intermediate Value Theorem.** If a real valued funtion f is continuous on the closed interval [a, b] and if N is a number strictly between f(a) and f(b), then there exists a number c in (a, b) such that f(c) = N.

**Mean Value Theorem.** If a real valued function f is continuous on the closed interval [a, b] and f is differentiable on the open interval (a, b) then there is a number c in (a, b) such that f'(c) = (f(b) - f(a))/(b-a).
We can use the Mean Value Theorem to prove that certain polynomials do not have more than one real root. (A root of a polynomial p(x) is a number c such that p(c) = 0.)

**Theorem.** The polynomial p(x) = x^{3} + x - 1 has exactly one real root.

**Proof.** The proof is in two parts.

**Part 1.** (Direct Existential Proof.) First we will prove p(x) has one real root. We appeal to the Intermediate Value Theorem with a = 0 and b = 1: p(0) = - 1 < 0 and p(1) = 1 > 0. Since 0 (N = 0) is between -1 (=p(0)) and 1 (=p(1)), we may conclude that there is a real number c, between 0 and 1, for which p(c) = 0.

**Part 2.** (Proof by Contradiction.) Now we will prove that p(x) has only one root. Assume to the contrary that p(x) has more than one root. Let's suppose two distinct roots c_{1} and c_{2}, so p(c_{1}) = p(c_{2}) = 0. Then by appealing to the Mean Value Theorem, there must be a number c between c_{1} and c_{2} for which p'(c) = (p(c_{2}) - p(c_{1}))/(c_{2} - c_{1}) = 0. But a direct calculation shows that p'(x) = 3x^{2} + 1, which can never be zero since x^{2} >= 0 for all real numbers x. The contradiction completes the proof.

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**Theorem.** If an object is traveling in a straight line with a differentiable position function s(t), where t denotes the time variable, for t between a and b, then there is a time t_{0}, between a and b where the instantanious velocity at t = t_{o} is equal to the average velocity over the entire path.

**Proof.** The velocity function is the derivative of the position function v(t) = s'(t). According to the Mean Value Theorem, there is a value t = t_{0} between a and b where v(t_{0}) = s'(t_{0}) = (s(b) - s(a))/(b-a) = average velocity over the path.

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- 4
^{25}6^{50}9^{25}- 1 is a composite number. (Constructive proof.) - There is a rational number that lies strictly between 19
^{ 100}- 1 and 19^{ 100}. (Constructive proof.) - sqrt(2) + sqrt(3) is an algebraic number. (See the section on Unwinding Defintions (Getting Started) for the definition of an algebraic number.) (Hint: First compute (sqrt(2) + sqrt(3))
^{2}.) (Constructive proof.) - Suppose f(x) is a real valued differentiable function (of real numbers) with f(1) = -1 and f(2) = -2, then there is a value of x = x
_{0}, between 1 and -2 such that the y-intercept of the tangent line to the curve at x = x_{0}is equal to f(t_{0}) + t_{0}. (Existential proof--use the Mean Value Theorem).

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