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How do you show that $$a^2 + 1$$ is never divisible by a $3 \mod 4$ integer (which is equivalent to showing that it has no $3 \mod 4$ prime factor) for any non-negative integer $a$ by analysing the arithmetic series representation of $a^2$, $1 + 3 + 5 + ... + (2a-1)$?

Måns Nilsson
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  • It doesn't use arithmetic continuation, but this is essentially trivial to show by considering even and odd cases separately. – Christian Sykes Aug 11 '17 at 08:45
  • If the variable expression had been in linear form instead of quadratic form, it would have been easy to explain why there are solutions, since then you could reach any integer you want. But it's not obvious why changing it to a perfect square skips all integers that are divisible by 3 mod 4 primes as $a$ is incremented. – Måns Nilsson Aug 11 '17 at 09:16

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Suppose $a \in \mathbb{Z}$ is such that $p\mid (a^2+1)$, where $p$ is prime, and $p \equiv -1 \pmod 4$.

Since $p\mid (a^2+1)$, it follows that $p \not\mid a$, hence $\gcd(a,p)=1$. \begin{align*} \text{Then}\;\;&p\mid (a^2+1)\\[4pt] \implies\;&a^2\equiv -1\pmod p\\[4pt] \implies\;&(a^2)^\frac{p-1}{2}\equiv -1\pmod p &&\text{[since $\small{\frac{p-1}{2}}$ is odd]}\\[4pt] \implies\;&a^{p-1}\equiv -1\pmod p\\[4pt] \implies\;&1\equiv -1\pmod p &&\text{[by Fermat's little Theorem]}\\[4pt] \implies\;&2\equiv 0\pmod p\\[4pt] \end{align*} contradiction.

Hmmm . . .

I didn't see the end of your question. It looks like you want a different kind of proof. In any case, the proof I gave above, using Fermat's little Theorem, is an easy way to prove the claim.

quasi
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    I was thinking about the series $$1 + 3 + 5 + ... + (2a - 1)$$ which is the arithmetic series for $a^2$. – Måns Nilsson Aug 11 '17 at 08:32
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    Was that given as a hint in your book? – quasi Aug 11 '17 at 08:33
  • I was looking for a proof based on logical and intuitive reasoning about the behavior of $1 + a^2$ over the non-negative integers of $a$, that explains why this diophantine expression cannot be expressed as the sum of exactly $3 mod 4$ copies of any one particular integer, and doesn't require a substitution from a difficult theorem like Fermat's little Theorem. But I can't say that this is wrong either, because it only relies on accepted theorems of mathematics. – Måns Nilsson Aug 11 '17 at 18:19
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Use congruences:

$a^2+1$ divisible by $p$ means $a^2\equiv -1\mod p$. In other words $-1$ is a square modulo $p$.

If you know the first supplementary law of the law of quadratic reciprocity, you know it is impossible if $p\equiv 3\mod 4$.

If you don't know it, here is a sketch:

If $-1\equiv a^2\mod p$, *Lil' Fermat says that $a^{p-1}\equiv 1\mod p$. However, as $p$ is odd, $$a^{p-1}=\bigl(a^2\bigr)^\tfrac{p-1}2=(-1)^\tfrac{p-1}2\equiv 1\mod p,$$ and this is impossible as $\dfrac{p-1}2$ is odd if $p\equiv 3\mod 4$.

J. W. Tanner
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Bernard
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Not using the series:

  • Case 1: $a$ is even, say, $a=2k$. Then $a^2+1 = 4k^2 + 1$.
  • Case 2: $a$ is odd, say, $a=2k+1$. Then $a^2+1 = 4k^2+4k+2$.

Using the series $a^2=1+3+5+...+(2a-5)+(2a-3)+(2a-1)$.

  • Note that consecutive terms add to multiples of $4$.
  • Therefore, $a^2$ is either $(1+3)+(5+7)+...+((2a-3)+(2a-1))$, which is a sum of multiples of 4, or $1 + (3+5)+(7+9)+...+((2a-3)+(2a-1))$, which is one more than a multiple of $4$.
  • in any case, the result follows for $a^2+1$

Explicitly in $\mathbb{Z}_4$:

  • $a$ is either 0, 1, 2 or 3. Therefore $a^2$ is either 0 or 1, so $a^2+1$ is 1 or 2.
Michael Hartley
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    Showing that $a^2+1$ is not congruent to $3$ mod 4 is not enough. The problem is to show that no prime factor of $a^2+1$ is congruent to $3$ mod 4. – quasi Aug 11 '17 at 08:57