The Student Room Group

Euclid Said!!!

Euclid's quote at the end of his posts says:


-"A person who can, within a year, solve - 92y² = 1 is a mathematician."-



I have been avoiding my revision like crazy but I have managed to solve the above equation. I reckon that x = 1151 and y = 120 will solve the above equation:


(you could of course change the sign of x or y as they are to be squared)


1151^2 - 92(120)^2 = 1

Does this mean I get to be a Mathematician? Not if I don't do some P4 revision, and soon. Anyone got any other solutions to this? Can there be any other solutions to this? (Except x = 1 and y = 0). All the best, Mensandan

Current solutions
x=1, y=0
x=-1, y=0
x=1151, y=120
x=1151, y=-120
x=-1151, y=120
x=-1151, y=-120
Reply 1
Another solution is
x = 2649601
y = 276240

edit; and another,
x = 14040770918401
y = 1463851560480
Reply 2
mensandan
Euclid's quote at the end of his posts says:


-"A person who can, within a year, solve - 92y² = 1 is a mathematician."-



I have been avoiding my revision like crazy but I have managed to solve the above equation. I reckon that x = 1151 and y = 120 will solve the above equation:


(you could of course change the sign of x or y as they are to be squared)


1151^2 - 92(120)^2 = 1

Does this mean I get to be a Mathematician? Not if I don't do some P4 revision, and soon. Anyone got any other solutions to this? (Except x = 1 and y = 0). All the best, Mensandan

Current solutions
x=1, y=0
x=-1, y=0
x=1151, y=120
x=1151, y=-120
x=-1151, y=120
x=-1151, y=-120


x=2649601 y=276240

Each successive solution is about 2300 times the previous
solution; they are every 8th partial fraction (x=numerator,
y=denominator) of the continued fraction for sqrt(92) =
[9, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, ...]

Once you have the smallest positive solution (x1,y1) you
don't need to "search" for the rest. You can obtain the nth positive
solution (xn,yn) by the formula.

(x1 + y1 sqrt(92))^n = xn + yn sqrt(92).

It's a Pell equation.

Euclid

(Look at first google site)
Reply 3
"mensandan" T. S. R. applaudes... A query of pure interest arises to the Mathematical mind... Would your respected T. S. R. colleagues be able to see YOUR complete solution.

Newton.
Reply 4
Newton
"mensandan" T. S. R. applaudes... A query of pure interest rises to the Mathematical mind... Would your respected T. S. R. colleagues be able to see YOUR complete solution.

Newton.

It might involve a certain website :biggrin:
Reply 5
JamesF
It might involve a certain website :biggrin:


LOL!

mensandan
I have been ...


Show your proof ... I'm just interested.

Euclid
Reply 6
Newton
"mensandan" T. S. R. applaudes... A query of pure interest rises to the Mathematical mind... Would your respected T. S. R. colleagues be able to see YOUR complete solution.

Newton.


I wish I could produce a complete solution, no such brains here I simply found the first non trivial solution using excel.
I generated a list of the first 65000 numbers (or so) in column A then produced a list of them (squared-1)/92 (in column B)
Finally in the next column i simply square rooted column B knowing that any integer answer would be necessarily a solution of the equation.

I have no idea how to solve this with a real proof but this Pell equation stuff sounds interesting, I can't open the link to the 'certain site' what is it? I did manage to open the first Pell equation site on google (wolfram) all looks jolly impressive if a little complex to me. Try my 'idiot' solution

Good luck to anyone with exams coming up soon, hope you are getting more work done than me, all the best, mensandan
Reply 7
y = 1
x = sqrt 93

:redface:
Reply 8
mik1a
y = 1
x = sqrt 93

:redface:


Integer solutions only my friend :biggrin:

Euclid
Reply 9
Euclid's judgment was not always infallible. :smile:
Reply 10
J.F.N
Euclid's judgment was not always infallible. :smile:


Judgement of what exactly? If you're refering to the qoute in my signature then let me tell you it is not in my judgement that one would take a year, it is Brahmagupta's.

Euclid
Reply 11
Euclid
Judgement of what exactly? If you're refering to the qoute in my signature then let me tell you it is not in my judgement that one would take a year, it is Brahmagupta's.

Euclid


Sarcasm, my friend. I'm talking about Euclid of Alexandria.
Reply 12
I would like to contribute what i know about Pell's equation

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Find integer solutions of the following equation, where a is an integer. (ALL numbers involved are integers)

X^2 - a.Y^2 = +/- 1

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Fact 1: a cannot be a perfect square, if there are any solutions to exist.

Proof: Let a be a perfect square ==> a=b^2, so

X^2 -b^2.Y^2 = +/- 1
X^2 -(b.Y)^2 = =/-1
(X-b.Y)(X+b.Y) = +/- 1
==> X-b.Y = +/- 1 AND X+b.Y = +/- 1 (in case when it equals 1)
Inconsistent, therefore no integer solutions.

when it equals -1:

==> X-b.Y=-1 AND X+b.Y=1 (in case when it equals -1)

==> X = b.Y-1 AND X = 1-b.Y
==> b.Y - 1 = 1 - b.Y
==> b.Y = 1
==> b=1 AND Y=1 OR b=-1 AND Y=-1
when b=1 a=1^2 = 1 (not a perfect square) Contradiction.
Hence no solutions in this case either.

Remark: This may have been simplified alot, but i made every step clear so

anyone can understand it.

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Fact 2: For X^2 - a.Y^2 = 1 , (1,0) is the trivial solution.

Proof: Substitute (1,0):

1^2 - a.0^2 = 1

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Fact 3: For X^2 - a.Y^2 = 1 , if (p,q) is one solution then so is:

X = [(p+q.√a)^n + (p-q.√a)^n]/2
Y = [(p+q.√a)^n - (p-q.√a)^n]/2.√a

Proof: If (p,q) is a solution then p^2 - a.q^2 = 1

==> (p^2 - a.q^2)^n = 1
==> X^2 - a.Y^2 = (p^2 - a.q^2)^n = 1
==> (X+Y.√a)(X-Y.√a) = (p+q.√a)^n(p-q.√a)^n
==> X+Y.√a = (p+q.√a)^n
X-Y.√a = (p-q.√a)^n

Adding gives:

2X = (p+q.√a)^n + (p-q.√a)^n
X = [(p+q.√a)^n + (p-q.√a)^n]/2

Subtracting gives:

Y.2√a = (p+q.√a)^n - (p-q.√a)^n
Y = [(p+q.√a)^n - (p-q.√a)^n]/2.√a

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Fact 4: X^2 - a.Y^2 = 1 is always solvable if a is non-square.

Proof: Based on the fact that you can always derive a continued fraction for √a.

--------------------------------------------------------------------------------

Fact 5: X^2 - a.Y^2 = -1 will have no solutions if a is a multiple of 4.

Proof: I dont know the formal proof.

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Reply 13
IntegralNeo
Fact 5: X^2 - a.Y^2 = -1 will have no solutions if a is a multiple of 4.

Proof: I dont know the formal proof.

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Work mod 4.