Math Is Fun Forum

Hi Faria.

Let m be the amount money and p the number of people. Then

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Now eliminate m and solve for p.

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Like a nanotube, but slightly larger.

Correct. BUT:

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Complex numbers are different.

So if you want to have

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one of the √(−1) is supposed to be i and the other −i. Similarly if

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one of the √(−1) is supposed to be i and the other two either both i or both −i. With a one–many relation you can't always tell.

Variation on I Timothy 6:10.

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Hello individ. Hope you don't mind me contributing to your thread. I just considered diophantines of the form

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This is solvable if and only if gcd(a,b) divides c. The general solution is

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where

and X=x, Y=y is a particular solution. See this post:

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To solve the following Diophantine equation

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Otherwise there is no solution. For example, 2x + 4y = 9 has no solution since the LHS is always even while the RHS is odd.

For example, 2x + 4y = 10  → x + 2y = 5. This step is optional but it helps by making the numbers you are working with smaller.

This can be done by trial and error. If the numbers involved are large, the Euclidean algorithm might be useful. (NB: A particular solution always exists. See following post.)

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Proof: Check first that

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Now suppose x=xʹ, y=yʹ is another solution. Then

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QED.