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- Thread starter Poopsilon
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- #2

Hurkyl

Staff Emeritus

Science Advisor

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Correct.that it has no integer solutions mod 5 to conclude the original equation has no integer solutions, correct?

If there was an integer solution, that would also be a solution to the reduced-modulo-5 version.

And since there isn't a solution to the reduced-modulo-5 version, there isn't an integer solution.

Any natural number. But via the Chinese Remainder Theorem, you only really need to deal with prime powers -- e.g. reducing modulo 6 tells you the exact same information as considering reducing modulo 2 and reducing modulo 3 separately. For example, there's the theorem:Also does this only work modulo a prime, or can I do this modulo any natural number?

A (polynomial) equation doesn't have a solution modulo 6 if and only if at least one of the following is true:

- The equation doesn't have a solution modulo 2
- The equation doesn't have a solution modulo 3

Incidentally, it's fairly common that 4 and 8 are more useful to consider than 2.

- #3

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Ah yes, that makes sense, thanks =].

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Would someone mind showing the steps to reducing that? (a^2-10b^2=2 to a^2 = 2 mod 5)

- #5

Deveno

Science Advisor

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Would someone mind showing the steps to reducing that? (a^2-10b^2=2 to a^2 = 2 mod 5)

to indicate a number modulo 5, i will write [n] instead of n, so

[13] = [3].

a

[a

[a

[a]

[a]

one can explicitly compute [a]

[0][0] = [0]

[1][1] = [1]

[2][2] = [4]

[3][3] = [9] = [4]

[4][4] = [16] = [1]

or, using an "old-fashioned method":

let a = a' + 5k

let b = b' + 5m

then a

= (a')

collecting all obvious multiples of 5, we get:

= a'

let n = 2a'k + 5k

a'

that is: a

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