Cauchy problem for PD equations with variable symbols

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E. lazy) way out. Obtaining a useful factorization is next to impossible, except when n is even. Then we get a difference of two squares (a vital factorization in Diophantine equations) like so: 7 = x2 − 2n = (x − 2m )(x + 2m ), where m = n/2. Then we can say that x − 2m and x + 2m , being factors of 7, must be −7, −1, 1, or 7; and further breakup into cases soon shows that there are no solutions (if we assume n is even). But that is about as much as the factorization method can tell us; it does not tell us where the actual solutions are and how many of them there are.

G. g. (mod pq)). Factorization alters the problem into the form (factor) × (factor) = (something nice), where the right-hand side could be a constant (the best possible result), a prime, a square, or something else that has a limited choice of factors. ). Now it is best to try elementary techniques first, as it may save a lot of dashing about in circles later. One may have abandoned these methods and tried to analyse the approximate equation x= 2n + 7 ≈ 2n/2 which can get into some serious number theory involving topics such as continued fractions, Pell’s equation, and recursion relations.

Well, it has at most n roots. Do we know anything about the roots of p? Well, p is a factor of f , so the roots of p are also roots of f . What are the roots of f ? There are none! ) f is always positive (in fact, it is always at least 1), and hence can have no roots. This means in turn that p can have no roots. What does it mean when a polynomial has no roots? It means that it never crosses 0, that is, it never changes sign. In other words, p is either always positive or always negative. This gives us two cases, but we can save a little bit of work by observing that one case implies the other.

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