# New PDF release: Basic concepts of differential algebra By Wuerfl A.

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Example text

In fact, in terms of u and v, we have 2 a 2 c−b 2 c+b 2 uv. Let p be a prime divisor of a/2. Then p2 divides uv; and since u and v are relatively prime, p2 must divide either u or v. It follows that u and v are pure square numbers: u s2 and v t 2 . In addition, s and t are also relatively prime and of different parity. With these, √ 2st. Putting everything together, we arrive we have a 2 uv at a b 2st, t 2 − s2 , c t 2 + s2 . We claim that this is the general form of a Pythagorean triple (with relatively prime components) parametrized by two integers s and t, where t > s, and s, t are relatively prime and of different parity.

We now use a standard trick in differential equations (due to Bernoulli) and multiply both sides by the integrating factor ekx to obtain [ekx F(x)] ekx [F (x) + kF(x)] k 2n+1 ekx f(x). Integrating both sides between 0 and 1: 1 k 2n+1 ekx f(x) dx 0 [ekx F(x)]10 ek F(1) − F(0) (1/b)(aF(1) − bF(0)). In other words, 1 bk 2n+1 ekx f(x) dx aF(1) − bF(0) 0 is an integer. , 0 < x < 1, we have 1 0 < bk 2n+1 ekx f(x) dx < 0 bk 2n+1 ek n! bkek (k 2 )n . n! For ﬁxed k, (k 2 )n /n! tends to zero as n → ∞. Thus, for n sufﬁciently large, the entire right-hand side can then be made < 1.

4 Let n ∈ N be a congruent number. By deﬁnition, there exists a right triangle with c2 rational side lengths a, b, c, and area n. We have a2 + b2 and ab 2n. Adding or subtracting twice the second equation c2 ± 4n. Setting x (c/2)2 , we from the ﬁrst, we get (a ± b)2 see that x, x + n, and x − n are squares of rational numbers. √ The converse √ of this statement is also √ x + n − x − n, b x + n + x − n, and true, √ since a c 2 x are the rational side lengths of a right triangle with area n, provided that x is a nonzero rational number with x, x + n, and x − n squares of rational numbers.