Non-linear Diophantine?

How many ordered pairs of integers satisfy the equation n 4 4 m = 3 n^4 - 4m = 3 ?


The answer is 0.

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1 solution

If we reduce the equation to modulus 4, we'll get that:

n 4 3 ( m o d 4 ) n^4 \equiv 3 (mod 4)

Now, spliting n 4 n^4 into two cases that cover all possibilities:

2 k 4 = 16 k 4 0 ( m o d 4 ) 2k^4 = 16k^4 \equiv 0 (mod 4)

( 2 k + 1 ) 4 = 16 k 4 + 32 k 3 + 24 k 2 + 8 k + 1 1 ( m o d 4 ) (2k+1)^4 = 16k^4 + 32k^3 + 24k^2 + 8k +1 \equiv 1 (mod 4)

So the division of a number on the form k 4 k^4 by 4 is either 0 or 1, meaning that we have a contradiction and the equation has no solutions.

It is easy to see, that we have basically the same situation at the generalised form of the question:

n 2 k ± 4 l m = 3 k , l Z , k > 0 n^{2k} ± 4lm = 3 \\ k, l \in \mathbb{Z} , k > 0

Zee Ell - 3 years, 11 months ago

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