
Number Theory * Introduction (1/22)
... Are there any (non-trivial) solutions in natural numbers to the equation a3 + b3 = c3? If so, are there only finitely many, or are the infinitely many? For any k > 2, are there any (non-trivial) solutions in natural numbers to the equation ak + bk = ck? If so, are there only finitely many, or are th ...
... Are there any (non-trivial) solutions in natural numbers to the equation a3 + b3 = c3? If so, are there only finitely many, or are the infinitely many? For any k > 2, are there any (non-trivial) solutions in natural numbers to the equation ak + bk = ck? If so, are there only finitely many, or are th ...
A GUIDE FOR MORTALS TO TAME CONGRUENCE THEORY Tame
... proves (1). Applying our knowledge of I = {1} shows (2). To prove (3) take a pair ha, bi ∈ ϑ − δ, and an element c ∈ O. Since h(I ∪ O, O) ⊆ O, the polynomial h(x, c) collapses ϑ into δ, thus hh(a, c), h(b, c)i ∈ δ. We can assume that ha, h(a, c)i 6∈ δ. But a = h(a, 1), thus hh(a, 1), h(a, c)i 6∈ δ a ...
... proves (1). Applying our knowledge of I = {1} shows (2). To prove (3) take a pair ha, bi ∈ ϑ − δ, and an element c ∈ O. Since h(I ∪ O, O) ⊆ O, the polynomial h(x, c) collapses ϑ into δ, thus hh(a, c), h(b, c)i ∈ δ. We can assume that ha, h(a, c)i 6∈ δ. But a = h(a, 1), thus hh(a, 1), h(a, c)i 6∈ δ a ...
EXAMPLE SHEET 3 1. Let A be a k-linear category, for a
... satisfies ei pej q “ δij . Prove that i“1 ei b ei P V b V is independent of the choice of the basis of V . 3. Let k be a field and Mn pkq the algebra of n ˆ n matrices with entries in k, and denote by OpMn pkqq be the free commutative algebra on the variables tXij : 1 ď i, j ď nu (ie the plynomial a ...
... satisfies ei pej q “ δij . Prove that i“1 ei b ei P V b V is independent of the choice of the basis of V . 3. Let k be a field and Mn pkq the algebra of n ˆ n matrices with entries in k, and denote by OpMn pkqq be the free commutative algebra on the variables tXij : 1 ď i, j ď nu (ie the plynomial a ...
2. 2 2 = 1 1 n i=1 n−1 i=1
... Given y > 0, define a sequence yi recursively by y1 = y and yi+1 = ri yi , andn let ai be the sequence obtained in this fashion when r 1 = 1. Then we have yi = ai y, and the problem amounts to finding a unique value of y such that X X X x = yi = ai y = ai · y . P Since b = ai is a sum of positive ...
... Given y > 0, define a sequence yi recursively by y1 = y and yi+1 = ri yi , andn let ai be the sequence obtained in this fashion when r 1 = 1. Then we have yi = ai y, and the problem amounts to finding a unique value of y such that X X X x = yi = ai y = ai · y . P Since b = ai is a sum of positive ...