- #1

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Also, what is the technical name for finding the mod n relations in an equation?

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In summary, the method to solving equations modulus a number can be found in a more complex form than a simpler form, and it is possible to find certain solutions using a Chinese Remainder Theorem. However, the case of n==1Mod 3 is not solvable, and is beyond the capabilities of the Chinese Remainder Theorem.

- #1

- 729

- 0

Also, what is the technical name for finding the mod n relations in an equation?

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- #2

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Now, if you want to solve it where each line in the linear system is modulo a different number, for instance x+3y=4 mod 5, and 2x-y=2 mod 6, then that method isn't going to work. And if it is a nonlinear equation, that is NP complete, so you'd best avoid trying unless you have a polynomial time algorithm for NP problems.

- #3

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The reference given, as you know, completely handles the case for n==2Mod3. This comes about from the form, where( n-2)/3 is an integer. It seems possible that the form was just found by chance. [tex]\frac{4}{n} =\frac{1}{n} +\frac{1}{(n-2)/3+1}+\frac{1}{(n^2+n)/3}[/tex]

Which, but not in the right form, simplifies to: [tex]\frac{4}{n}=\frac{1}{n}+\frac{3}{n+1}+\frac{3}{n^2+2}[/tex]

This is a more complex form harder to find than a simpler form, which may have been a starting point for the above, [tex]\frac{2}{n} = \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n^2+n}[/tex]

This uses a splitting method and if length is not a problem, we could replace 1/(n+1) with 1/(n+2) + 1/(n+2)(n+3) Thus the length of the Egyptian fraction could be increased and increased, but in the case under consideration we are only interested in three term Egyptian fractions.

Maybe you want to study Egyptian fractions further, I am not exactly sure of your questions. The standard method on solving multiple congruences is the Chinese Remainder Theorem.

Now on that simple form above we can find things like 1/5=1/10+1/11+1/110. So that 2/p can always be found in three terms of an Egyptian fraction.

However, in the specialized case of 4/n dealing with 3 Egyptian fractions, Mordel has shown that certain forms will not work like the case of n==2 Mod 3. In fact, Mordel shows that n==1 Mod 3 will not work the same way so we can not rely on a similar polynominal identity for all cases.

Which, but not in the right form, simplifies to: [tex]\frac{4}{n}=\frac{1}{n}+\frac{3}{n+1}+\frac{3}{n^2+2}[/tex]

This is a more complex form harder to find than a simpler form, which may have been a starting point for the above, [tex]\frac{2}{n} = \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n^2+n}[/tex]

This uses a splitting method and if length is not a problem, we could replace 1/(n+1) with 1/(n+2) + 1/(n+2)(n+3) Thus the length of the Egyptian fraction could be increased and increased, but in the case under consideration we are only interested in three term Egyptian fractions.

Maybe you want to study Egyptian fractions further, I am not exactly sure of your questions. The standard method on solving multiple congruences is the Chinese Remainder Theorem.

Now on that simple form above we can find things like 1/5=1/10+1/11+1/110. So that 2/p can always be found in three terms of an Egyptian fraction.

However, in the specialized case of 4/n dealing with 3 Egyptian fractions, Mordel has shown that certain forms will not work like the case of n==2 Mod 3. In fact, Mordel shows that n==1 Mod 3 will not work the same way so we can not rely on a similar polynominal identity for all cases.

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- #4

- 1,059

- 1

[tex]\frac{4}{n}=\frac{1}{n}+\frac{3}{n+1}+\frac{3}{n^2 +2}[/tex]

The above was a mistake, and should be: [tex]\frac{4}{n}=\frac{1}{n}+\frac{3}{n+1}+\frac{3}{n^2 +n}[/tex]

Here is my improvised form: [tex]\frac{4}{n} = \frac{1}{n/3}+\frac{1}{n+1}+\frac{1}{n^2+n}[/tex]

Which will work as long as 3 divides n, for example= 4/15=1/5+1/16+1/240.

Now the only thing we would need to compete the problem would be to solve for n==1Mod 3, which Mordel assures us is not possible. However with a minus sign, we would have: [tex]\frac{4}{n} = \frac{1}{n} +\frac{1}{(n-1)/3}-\frac{1}{(n^2-n)/3}[/tex]

The above was a mistake, and should be: [tex]\frac{4}{n}=\frac{1}{n}+\frac{3}{n+1}+\frac{3}{n^2 +n}[/tex]

Here is my improvised form: [tex]\frac{4}{n} = \frac{1}{n/3}+\frac{1}{n+1}+\frac{1}{n^2+n}[/tex]

Which will work as long as 3 divides n, for example= 4/15=1/5+1/16+1/240.

Now the only thing we would need to compete the problem would be to solve for n==1Mod 3, which Mordel assures us is not possible. However with a minus sign, we would have: [tex]\frac{4}{n} = \frac{1}{n} +\frac{1}{(n-1)/3}-\frac{1}{(n^2-n)/3}[/tex]

Last edited:

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