I don't know where this goes but seems algebraic/equation-like...

[randomperson]

Find all pairs of odd integers m and n which satisfy the following equation:
m+128n=3mn

 

[cordoba]

Here's a hint:

Any odd integer, n, can be written as:

\(\displaystyle n = 2n_1 + 1, n_1 \in Z\)

Try using this expression in your equation.

 

[lookagain]

\(\displaystyle 8a + 262b - 130 = 12ab \implies > > 4a < < + 131b - 65 = 6ab.\)

. . .

What do I mean? Well 8a is even as is 6ab. So 131b - 65 cannot be odd.
Do you see why? But that means 131b IS odd. So that means b is odd.
So we can revise our equation once again.

JeffM,

you must have meant to have typed "Well 4a is even as is 6ab,"
as you were referring to the second (implied) equation.

I'm thinking you accidentally read across two parts of the
different forms of the equation when you typed out the
second equation and morphed them together.


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


Edit:

... But that means 131b IS odd. So that means b is odd.
So we can revise our equation once again.

Let b = 2c - 1.

Off you go. Presumably you will eventually get somewhere where
only a few possibilities are left.

When I tried that substitution, it only complicated the equation more for me:


4a + 131(2c - 1) - 65 = 6a(2c - 1) ==>

4a + 262c - 131 - 65 = 12ac - 6a ==>

4a + 262c - 196 = 12ac - 6a ==>

10a + 262c = 12ac + 196 ==>

5a + 131c = 6ac + 98


As the right-hand side is even, so is the left-hand side.

But then we're stuck with either both a and c are odd,
or both a and c are even.

And we're left with more questions.

 

[lookagain]

Integer solutions, including evens and negatives (m,n):
0,0
32,-1
40,-5
42,-21
43,43
44,11
48,3
64,1

Yes, but what method was used to generate this list of solutions?

 

[lookagain]

It is even possible to "turn" this style of equation
(where solutions = odds)into a quadratic, like this:
n = 2a + 1

> > m = 2a + 2b + 1 < <

Denis,

why this expression for m?

Why not m = 2b + 1, allowing for a and b to equal each other
at sometime (such as m = n= 43)?


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If > > m < 0 > 0 < < , then 3mn > 0 and m + 128n < 0 so ...

JeffM,

what does this mean? Is it a typo?

_________________________________



Edit # 1:

I am going to assume that m and n must be positive integers.

\(\displaystyle m + 128n = 3mn \implies \)

\(\displaystyle 128n = 3mn - m = m(3n - 1) \implies \)

\(\displaystyle m = \dfrac{128n}{3n - 1} \implies \)

\(\displaystyle m = n * \dfrac{2^7}{3n - 1} \ implies \)

\(\displaystyle dfrac{2^7}{3n - 1}\ is\ an\ integer \implies\)

JeffM,

I don't see the justification in these steps (at the end).

At this stage, as you stated, m and n are assumed to be positive integers,

so how can \(\displaystyle \dfrac{2^7}{3n - 1} \ \ be \ \ an \ \ integer \ \ with \ \ certainty ?\)


What if (3n - 1) and n were to share at least a 2 as a factor, as in

\(\displaystyle \dfrac{(2k_1)(2^7)}{(2^8)k_2}, \ \ \ where \ \ k_1 \ \ and \ \ k_2 \ \ are \ \ integers?\)


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Edit # 2:

JeffM,

I feel your post after this one has explained more to me.

But still I am asking, what does m < 0 > 0 mean?

 

[Colacanth]

Would this work?

 

[Colacanth]

Thank you.

I owe this to my eight fins, unhealthy consumption of cola, and completing rectangles.