diophantine equation

[bugzpodder]

so the question boils down to solve for all positive integer n,m such that:
3n*(n+1)/2=2*m*(m+1)

[sql_lall]

Interesting

lets see....from the series 1->k rule we have:
1) 3n(n+1)/2 = 3(1+2+3+4+...+n)
2) 2m(m+1) = 4(1+2+3+4+..+m)
Obviously this means n>m as both are >0
Just one thing to think about.
this leads to:
1+2+3+...+m = ([m+1]+[m+2]+...+n)*3
=> 6 is a factor of m(m+1)
=> m^2+m-6k=0
=> m= [-1 + sqrt(1+24k^2)]/2
=> narrows down possibilies for m (though not all work, like when k=1)



Also, using quadratic formula:
3n^2+3m=4m^2+4m
=> m=(-1+sqrt(3n^2+3n+1))/2
=> 3n^2+3n+1=d^2, where d>1

Just a few ideas..no answers though

[bugzpodder]

...

[sql_lall]

Some solutions include:

n,m=
7,6
104,90
1455,1260
20272,17556
282359,244530
3932760,3405870
54776287,47437656

Just thought u'd like to know.
All the right hand side are multiples of 6.


P.S. Thanks go to someone else who programmed to find these.

[NotLKH]

Originally posted by sql_lall
Some solutions include:

n,m=
7,6
104,90
1455,1260
20272,17556
282359,244530
3932760,3405870
54776287,47437656

Just thought u'd like to know.
All the right hand side are multiples of 6.


P.S. Thanks go to someone else who programmed to find these.

Also, if you say:

M₀ = 0
M₁ = 6
M₂ = 90

then M_(k+1) = 14*M_k - M_(k-1) + 6



PS:
I believe

762935264, 660721320
10626317415, 9202660830
148005508552, 128176530306
2061450802319, 1785268763460
28712305723920, 24865586158140
399910829332567, 346332937450506
....


also works,

And, there's an interesting event if you check N_(k+1) out.

[NotLKH]

Looks like the semi-generic sequence for N_(k+1) and M_(k+1) is the following:

Given a*N*(N+1) = b*M*(M+1), where
b = a + 1, then:

N_(k+1) = 2*N₁*N_k - N_(k-1) + M₁
M_(k+1) = 2*N₁*M_k - M_(k-1) + M₁

where:
N₀ = 0 , M₀ = 0
N₁ = 2a+1 , M₁ = 2a



-Lou


hmmm,
now what's the generic iterative sequence for N_(k+1) and M_(k+1)
when, given a*N*(N+1) = b*M*(M+1),
b = a + t, t > 0?

[NotLKH]

hmm,

I just wanted to investigate aX(X+1) = bY(Y+1), specifically
I put a progie together to return the first nonzero X:Y as a and b varies from 1 to 25, testing all values X:Y from 1 to 3000.

{returns a tab deliminated table to the clipboard, where a varies from left to right, and b varies from top to bottom}

Obviously, its symetric across the a = b diagonal.
And, I noticed that when a is a square and b is a square, when a <> b, then
either there are no first nonzero X₁:Y₁ pairs, or else they are > 3000.

Secondly, of course, when a and b are non-relative prime, they match their reduced values cell entry:

ie... cell(3, 6) = cell(1,2)

So, for all to play with, attached is a zipped pdf displaying the results as a, b varies from 1 to 25.




-Lou

[NotLKH]

And, I was wondering,
Can anyone figure out the iterative sequence for X, Y from the following:

X(X+1) = 7Y(Y+1)

X₀ = 0, Y₀ = 0, and the pairs
X_k, Y_k, as k = 0 thru 6 are:

0 0
6 2
14 5
104 39
231 87
1665 629
3689 1394


-Lou

[NotLKH]

And, if anyone wants, Here's a progie that tests X, Y over a range of 1 thru 5000, such that it returns the X, Y that satisfies:

aX² + bX + c = dY² + eY + f

{Just as long as you don't overflow it, and always use numeric values}


-Lou

[bugzpodder]

i'll have some details tomorrow

[NotLKH]

Originally posted by NotLKH
Also, if you say:

M₀ = 0
M₁ = 6
M₂ = 90

then M_(k+1) = 14*M_k - M_(k-1) + 6



...

Hmm, I just noticed N_k = (1/6)*(7*M_k - M_(k-1))

[NotLKH]

Originally posted by bugzpodder
i'll have some details tomorrow

Shouldn't be too much longer.
I can't get past the Iterative solution for N_k and M_k!

I'll play some more, see if I can get a power series based on 14^k working, but it seems difficult to generalize.


-Lou

[NotLKH]

Hmmmm,

Is this the form of an answer you want?

X_n+1 = P*X_n + Q*Y_n + K
Y_n+1 = R*X_n + S*Y_n + L

I was starting to think Iterative sequences weren't what you wanted.


-Lou

[bugzpodder]

(m_1,n_1)=(6,7)
(m_2,n_2)=(90,104)

and m_k=14m_(k-1)-m_(k-2)+6
n_k=14m_(k-1)-m_(k-2)+6

and of course, (m_k,n_k) are solutions to the equation

[NotLKH]

Originally posted by NotLKH
Also, if you say:

M₀ = 0
M₁ = 6
M₂ = 90

then M_(k+1) = 14*M_k - M_(k-1) + 6


And, there's an interesting event if you check N_(k+1) out.

Cool. I was right.

By the way, you might like to check this site out:

http://www.alpertron.com.ar/QUAD.HTM

It supposedly solves Diophantine equations.

-Lou