| Q. p�,q�,r� form an arithmetic progression, with p,q,r all +ve integers
and p > q > r. Characterize the possible solutions.
A.
p = �n(e� + 2eh -h�)
q = �n(e� + h�)
r = �n|e� -2eh -h�|
with n,e,h all +ve integers, e > h both odd and coprime.
Proof:
First, check the solution above. p�,q�,r� is an arithmetic progression,
with p>q>r>0 as required.
Secondly, no arithmetic progression can have two distinct characterizations,
n,e,h & n',e',h' say. To see this, show that (p,q)=n, and then note that
one of the pair {(p+r)/n,(p-r)/n} is 0 mod 8, the other is 4 mod 8. The
former is e�-h�, and combined with the knowledge that e�+h� = 2q/n, we're home.
But how do we get the magic formula? And why should every valid a.p. take
this form?
To begin, the a.p. constraint can be written as:
p� + r� = 2.q�
Now, let's generalize the problem slightly to
p� + r� = q� + s�
=> (p+q)(p-q) = (s+r)(s-r)
where we have p,q,r,s +ve integers as before, and p>q,p>s (so s>r)
Next, we need the following...
Lemma: If a,b,c & d are positive integers such that ab=cd, then we can find
unique positive integers j, e,f,g & h such that:
(e,h) = 1
(f,g) = 1
a = jef
b = jgh
c = jeg
d = jfh
Proof of Lemma:
Given +ve integers a,b,c,d such that ab = cd, let j = (a,b,c,d)
so that a = jA, b=jB, c= jC, d =jD. Thus AB = CD.
Condidering all prime factors of A,B,C & D in turn, we see that:
A = (A,C).(A,D)
B = (B,C).(B,D)
C = (A,C).(B,C)
D = (A,D).(B,D)
furthermore ((A,C),(B,D)) = 1
and ((A,D),(B,C)) = 1
Indeed, any +ve integers, e,f,g,h,j where (e,h) = (f,g) = 1 generate a
general solution, with
a = jef
b = jgh
c = jeg
d = jfh
Now, in our case, a,b,c,d are p+q,p-q,s+r,s-r respectively. So:
p = �j(ef + gh)
q = �j(ef - gh)
r = �j(eg - fh)
s = �j(eg + fh)
However, we want to constrain things still further, to say q = s.
j <> 0, so:
ef - gh = eg + fh
=> e(f-g) = h(f+g)
(e,h) = 1, so ke = f+g for some k, similarly lh = f-g for some l,
ehk = ehl => k=l.
f = �k(e+h)
g = �k(e-h)
Since (f,g) = 1, k = 1 or 2.
If k = 1: e=f+g, h=f-g
e=h(mod2) => e=h=1(mod2) => f<>g(mod2).
Given any e,h odd with (e,h)=1, (f,g) = (�(e+h),�(e-h)) = (�(e+h),e)
= (since e odd) ((e+h),e) = (e,h) = 1.
4p = j(e� +2eh -h�)
4q = j(e� +h�)
4r = j|e� -2eh -h�|
For p� to be > r�, need e>h.
If k = 2: f=e+h, g=e-h
f=g(mod2) => f=g=1(mod2) => e<>h(mod2).
As above, any f,g odd with (f,g)=1 ensure that (e,h)=1.
4p = j(f� +2fg -g�)
4q = j(f� +g�)
4r = j|g� +2fg -f�|
For p� to be > r�, need f>g.
In either case, to ensure that p,q,r are all integer, we need j even.
So, basically the two values of k generate the same solutions:
p = �n(e� + 2eh -h�)
q = �n(e� + h�)
r = �n|e� -2eh -h�|
with n,e & h positive integers, e > h both odd and coprime.
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