Prove that there are always three naturals $a, b, c \in (n^2, (n + 1)^2): c \mid (a^2 + b^2); \forall n \in \mathbb Z^+, n > 1$.

Question

Prove that there are always three pairwise distinct naturals $$\large a, b, c \in \left(n^2, (n + 1)^2\right): c \mid (a^2 + b^2); \forall n \in \mathbb Z^+, n > 1$$

Here's an attempt of mine. Let $$a - x = b - y = c - z = n^2\implies x, y, z \in (0, 2n + 1)$$

It is necessitated to be sufficient to show that

$$(n^2 + z) \mid [(n^2 + x)^2 + (n^2 + y)^2] \iff (n^2 + z) \mid [(x - z)^2 + (y - z)^2]$$

But it can't think of anything anymore. Perhaps I could use apagogical arguments but in the cases of three changeable positive integers, that might not be disproven easily.

Answer 1

There isn't much to this other than trial and error, formulating a guess, and verifying that it works. So it's a good question for you to hone your skills on. Avoid peeking too much.

Hint: Play with small cases, find a pattern.

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For $ n = 2$, the only possibilities are $ 5 \mid 6^2 + 7^2$ and $5 \mid 6^2 + 8^2$.

If you really want a (potential) value:

$ c = n^2 + 1$ works. (Or we would really like it given the above. Of course, it could be $ c = n^2 + n -1$ too)

Continue with other cases given the above guess.

For $n=3$, we have $ 10 \mid 11^2 + 13^2$, $ 10 \mid 12^2 + 14 ^2 $.
Note: We can quickly verify that for $n=3$, $c = n^2 + n - 1=11$ would not work, so our previous guess about the value of $c$ might be accurate.
Note: The only other solution is $ 13 \mid 10^2 + 15^2$. This "seems" out of place, so we will ignore it for now.

If you really want another (potential) value:

Notice that $ 6 = 2^2 + 2, 12 = 3^2 + 3$. So we hope that $n^2 + n$ will work.

And if you must have the answer, the above cases - $5\mid 6^2 + 8^2, 10 \mid 12^2 + 14^2$ - yield the family:

$(a,b,c) = (n^2 + n, n^2+n+2, n^2 + 1)$.
$a^2 + b^2 = c\times 2(n^2+2n+2)$

Of course, there could be other families to consider too. For example, $ 5 \mid 6^2 + 7^2, 10 \mid 11^2 + 13^2 $ yields the family

$(a,b,c) = (n^2 + 2, n^2+n+1, n^2 + 1)$
$a^2+b^2 = c \times (2n^2 + 2n + 5)$
Note that this is the family that your original working suggests, which occurs when $ z = 1, y-z = 1, x-z = n$.