New results regarding the full solution of the diophantine equationx2+2k=yn in positive integers are obtained. These support a previous conjecture, without providing a complete proof.

The last twenty years has seen considerable and fruitful research in the field of nonsmooth boundary value problems (BVP’s) for partial differential equations. The objective is to understand the behavior and properties of solutions to either variable coefficient equations with minimal regularity assumptions on the coefficients or to linear constant coefficient equations in domains with nonsmoot...

Results When HR IgE FHS were categorized by eczema severity, those with severe AD were more likely to have HR IgE to cow’s milk (x2 11.3, P < 0.001), egg (x2 6.5, P < 0.05) and peanut (x2 25.4, P < 0.0001) compared to those with mild to moderate AD. Those ≤ 2 years with severe AD were more likely to have multiple HR IgE FHS compared to those with mild to moderate AD for all combinations: cow’s ...

2.1 Definition: Lipschitz Function A real valued function f : D ⊆ R → R is called Lipschitz continuous or is said to satisfy a Lipschitz condition if there exists a constant K ≥ 0 such that for all x1, x2 in D |f(x1)− f(x2)| ≤ K|x1 − x2| (3) The inequality is (trivially) satisfied if x1 = x2. Otherwise, for x1 6= x2, one can equivalently define a function to be Lipschitz if and only if |f(x1)−f...

A multidimensional version of the first Darboux problem is considered for a model second order degenerating hyperbolic equation. Using the technique of functional spaces with a negative norm, the correct formulation of this problem in the Sobolev weighted space is proved. In a space of variables x1, x2, t let us consider a second-order degenerating hyperbolic equation of the type Lu ≡ utt − |x2...

Consider the following example (typical in college algebra): 1 x2 + 1 x2 + 1 + 1 x2 + x + 1 x2 + x+ 1 = 4x + 6x + 8x + 6x + 3x+ 1 x2(x+ 1) (x2 + 1) (x2 + x+ 1) . Now let’s assume that all the coefficients in the above are from the binary field F2 = Z2 = {0, 1}. The result becomes much cleaner: 1 x2 + 1 x2 + 1 + 1 x2 + x + 1 x2 + x+ 1 = 1 (x2 + x)(x4 + x) . After a brief introduction to finite f...