Prove that for all natural numbers $n$, there exist distinct integers $x, y, z$ for which,


How to prove this using mathematical induction?

Some context:

A related question asks for solutions of $x^2 + y^2 + z^2 = 3^{10}$ where the asker first tries to express $3^1$ and $3^2$ as sums of three squares and then combine these to construct a representation of $3^3$, and so on. However, none of the answers seem to use this approach.

Legendre's three square theorem states that $n \in \mathbb{N}$ can be expressed as a sum of three squares if and only if $n$ is not of the form $4^a(8b+7)$, and $14^k$ clearly isn't. However, it does not guarantee that $x,y,z$ are distinct and applying it is not (at least not directly) a proof by induction.

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    $\begingroup$ @mechanodroid While I appreciate your efforts to improve a question that was of very poor quality, I don't think that it is appropriate to put words into the mouth of the person who originally asked the question. Indeed, were you still of low enough reputation that your edits needed to go through the queue, I would have rejected your edit on the ground that is "Clearly conflicts with the author's intent." $\endgroup$ – Xander Henderson Jul 9 '18 at 15:05
  • $\begingroup$ @XanderHenderson I'm aware of that, but have a look at this meta thread. Some of the answers and comments seem to support the view that adding context to other users' questions is appropriate if the question was of some interest in the first place (this one was even a HNQ at some point). $\endgroup$ – mechanodroid Jul 9 '18 at 17:15
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    $\begingroup$ @mechanodroid From the most upvoted answer in that thread: "However, one should make sure to respect the author's intent." You have added text to the question in the first person that is not clearly aligned with the original author's intent. There are no comments here that suggest that the author had read the linked question, nor that they are familiar with the Legendre's three square theorem. You do not know that this is in line with the original author's intent. Again, please don't put words in other people's mouths. $\endgroup$ – Xander Henderson Jul 9 '18 at 18:14
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    $\begingroup$ @mechanodroid to avoid the objections raised it can help to say the same thing but adopt a different style. I think it can be a problem to assert that OP is aware if a result when this is not a given. This does provide context but maybe it is a 'wrong' context. $\endgroup$ – quid Jul 9 '18 at 19:00

For $n=1$ we have $$14=1^2+2^2+3^2$$ For $n=2$ we have $$14^2=196=4^2+6^2+12^2$$ If $14^n=x^2+y^2+z^2$ then $$14^{n+2}=14^2(x^2+y^2+z^2)=(14x)^2+(14y)^2+(14z)^2$$ so you can do induction on even and odd numbers separately.

  • $\begingroup$ Also, 14⁰ = 1² + 0² + 0², if you count the natural numbers from 0. $\endgroup$ – Davislor Jun 30 '18 at 17:03
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    $\begingroup$ @Davislor $x,y,z$ are supposed to be distinct. $\endgroup$ – mechanodroid Jun 30 '18 at 20:04
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    $\begingroup$ Ah. Then it's not true for $n = 0$. The problem must be counting from 1. $\endgroup$ – Davislor Jun 30 '18 at 20:34

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