Divisibility strikes again

Number Theory Level 2

Find the number of integers $m$ for which $n^{4} + m$ is composite for all natural numbers $n$ .

infinite 248 100 65

1 solution

Eddie The Head
Apr 15, 2014

Let us take $m = 4k^{4}$ ,where $k$ is any natural number. So $n^{4} + m =n^{4} + 4k^{4}$ $((n^{2}+(2k)^{2})^{2} - (2nk)^{2}$

I think you mean "take $m = 4k^4$ ".

Also, can you explain why the numbers will be composite?

Calvin Lin Staff - 7 years, 1 month ago

Sir, from the last line we can write $(n^{2} + 4k^{2} - 2nk)(n^{2} + 4k^{2} + 2nk)$ and hence composite

Eddie The Head - 7 years, 1 month ago

Yes, that is missing from what you are trying to say. Currently, all I see is $n^4 + 4k^4$ , and then another line of $( (n^2 + *2k)^2)^2 - (2nk)^2$ . It is not clear what link you are trying to show, and the solution is not easily understood. If you intended the statements as a hint, state that they are a hint and the reader is expected to work it out from there.

Note that you also need to check that none of those factors are 1 or -1.

Calvin Lin Staff - 7 years, 1 month ago

Divisibility strikes again

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