Question

(Theory number)
Find all prime numbers p, such that
7p = 8x^2 - 1
P^2 = 2y^2 - 1
with x, y are integers.

Answer 1

By using trials i got p = 41. But im not sure if many values other of p.

Answer 2

7p = 8x^2 - 1
p^2 = 2y^2 - 1

subtract and get
p(p-7) = 2(y-2x)(y+2x)

Answer 3

Ok. So, what next ?

Answer 4

not sure, still working on it..

Answer 5

7p = 8x^2 - 1
p^2 = 2y^2 - 1
according the first eq, we know that p is odd prime.

Subtract, get
p(p-7) = 2(y-2x)(y+2x)

p-7 = even = 2(y - 2x) = 2y - 4x
and
p = y + 2x
subtract again, get
-7 = y - 6x or
y = 6x - 7

here we can see the relation betwen x and y
:)

Answer 6

Ah, yes. Just subtitute y = 6x - 7 into
p^2 = 2y^2 - 1, it become
P^2 = 2(6x - 7)^2 - 1
p^2 = 72x^2 - 84x + 97

From the equation.,
7p = 8x^2 - 1
Square both side, get
49p^2 = 64x^4 - 16x^2 + 1
49(72x^2 - 84x + 97) = 64x^4 - 16x^2 + 1
then solve for x ...

Answer 7

Ask wolfram only x = 1 and x = 6, as integers. Check one by one only x = 6 causes p be a prime.

http://m.wolframalpha.com/input/?i=64x%5E4+-+3544x%5E2+%2B+8232x+-+4752+%3D+0&x=0&y=0