1 Basic Notions - Caltech Mathematics Department

So the rational solutions of (2) are obtained by setting
X =
1 − t2 2t
, Y = ,
1+t2 1+t2 with t = ±1, 0, together with (±1, 0) and (0, ±1).
,u,v∈Z. Then
Write t = u
v
X = u2 − v2 u2 2uv
, Y =
+ v2 u2 + v2 It follows that the non-zero solutions in Z of (1) are given by
with
x = u 2 − v 2 ,y=2uv, z = u 2 + v 2
u = ±v, u,v = 0
To get primitive solutions, it is convenient to put
m = u + v, n = u − v
x =(u + v)(u − v) =mn, y = m2 − n2 ,z=
2
m2 + n2 2
For primitive solutions, take m, n odd ≥ 1, m>n. Check that these are all
the primitive solutions.
5 Linear Equations
Basic problem: Fixa1,...,an ∈ Z, n>0. Consider the equation:
a1x1 + ...anxn = a · x = m, (*)
where a =(a1,...,an) andx =(x1,...,xn). Determine if (*) can be solved
in integers. If so, determine all the solutions.
These are the simplest Diophantine Equations.
Earlier, we proved that, given a1,...,an ∈ Z, not all zero, ∃! positive
integer d, thegreatest common devisor, such that we can solve
a1x1 + ...anxn = m
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