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• Reflection on the conjecture that if q is the next prime after p, where p is a positive integer prime, and if q - p 2, then sqrt( [ p^2 + q^2 ] /2 -1 ) is
Message 1 of 1 , Jul 6, 2006
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Reflection on the conjecture

that if q is the next prime after p,

where p is a positive integer prime,

and if q - p > 2,

then

sqrt( [ p^2 + q^2 ] /2 -1 ) is irrational.

Case 1: q - p = 4.

Then set

p = 6 D + 1

q = 6 D + 5

p^2 + q^2 = [ 36 D^2 + 12 D + 1 ] + [36 D^2 + 60 D + 25 ]
= 72 D^2 + 72 D + 26

[p^2 + q^2 ]/2 = 36 D^2 + 36 D + 13

[p^2 + q^2 ]/2 - 1 = 36 D^2 + 36 D + 12
= 12 [ 3 D^2 + 3 D + 1 ]

= 4( 3 [ 3 D^2 + 3 D + 1) )

To be a counter example to the conjecture,

z2 = 4(3 [ 3 D^2 + 3 D + 1] ) would have to be a perfect square,

which is impossible,

because z2 is a multiple of 3, but not a multiple of 9.

Case q - p = 6

Set p = 6 D + t, where t = either 1 or -1.

Then q = 6(D+1) + t

p^2 + q^2 = [36 D^2 + 12 t D + t^2] + [36 D^2 + 72 D + 36 + 12 t D + 12 t +
t^2 ]

= 72 D^2 + [ 24 t +72 ]D + [12 t + 2 t^2 ]

[ p^2 + q^2]/2 = 36 D^2 + [ 12 t + 36 ] D + [6 t + t^2 ]

= 36 D^2 + [12 t + 36 ] D + [6 t + 1 ]

[ p^2 + q^2 ] /2 - 1 = 36 D^2 + [12 t + 36 ] D + 6 t

To be a counter example to the conjecture,

z2 = 36 D^2 + [ 12 t + 36 ] D + 6 t would need to be a perfect square.

Case t = -1.

z2 = 36 D^2 + 24 D - 6 = 3 ( 12 D^2 + 8 D - 2)

cannot be a perfect square because

z2 is divisible by 2 and not divisible by 4.

Case t = 1

z2 = 36 D^2 + 48 D + 6 = 6(6 D^2 + 8 D + 1)

cannot be a perfect square because

z2 is divisible by 6 and not divisible by 36.

case q - p = 6 M with M > 0.

p = 6 D + t with t^2 = 1.

q = 6 D + 6 M + t = 6 (D + M) + t

p^2 + q^2 = 36 D^2 + 12 t D + t^2
+36 D^2 + 72 M D + 36 M^2
+ 12 t D + 12 t M
+ t^2

z2 = [p^2 + q^2]/2 - 1 = 36 D^2 + 24 t D + 72 MD + 18 M^2 + 6 t M

= 3 (12 D^2 + 8 t D + 24 M D + 6 M^2 + 2t M)

Perhaps this case can lead to a counter example.

It's not clear to me yet.
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