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Prove that there exist infinitely many integers n such that $n, n + 1, n + 2$ are each the sum of the squares of two integers:

Solution: First solution: Let a be an even integer such that $a^2 + 1$ is not prime. (For example, choose a ≡ 2 (mod 5), so that $a^2 + 1$ is divisible by 5.) Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$, and setting x = (r + s)/2, b = (r − s)/2. Finally, put n = $x^2 − 1$, so that $n = a^2 + b^2$, $n + 1 = x^2$, n + 2 = $x^2 + 1$

I got lost in the following part:

Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$. What is rs?. Can someone expand this proof, so it looks clearer?

User8976
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ElChavoDelOcho
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  • That's a somewhat weird proof $a^2 + 1$ can be prime by setting $s=1$ and $r=a^2 + 1$. And we can either have $a^2 +1$ be odd or $a^2 + 1$ be divisible by $8$ we just need $a^2 +1 = r*s = (\frac {r+s}2 + [r-\frac{r+s}2])([\frac{r+s}2 - s]-\frac{r+s}2)$ so that $r$ and $s$ are both even or both odd. – fleablood Dec 18 '17 at 02:50
  • This problem has already been addressed there. Do not ask repeated questions. https://math.stackexchange.com/questions/2170494/infinitely-many-n-such-that-n-n1-n2-are-each-the-sum-of-two-perfect-squa – individ Dec 18 '17 at 04:34

3 Answers3

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If $a^2 +1$ is not prime, then it can be written as a product:

$$a^2 + 1 = rs,$$ where $r \geq s \gt 1$.

$r$ and $s$ are simply these factors here.

Matt
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  • how can you expand this further by explaining why we can do the next setting? – ElChavoDelOcho Dec 17 '17 at 23:56
  • Well, I hope you agree that we can define $x$ in the way that you have written in your question, and also $b$. I will leave it to you to work out why $x$ and $b$ are integers (this is important; if they weren't integers then you'd be stuffed). You can then define the integer $n$ in the way in which you have written, giving you the desired trio $n, n+1, n+2$, completing the proof. It's an elegant proof - particularly because proofs like this are (in my purely undergraduate experience) often done by contradiction, and this isn't. – Matt Dec 18 '17 at 00:00
  • Maybe $x=\frac{r+s}{2}, b=\frac{r-s}{2}$ is the key here? –  Dec 18 '17 at 00:01
  • It's certainly a key part of the proof, although (without trying to sound condescending) I currently am not seeing what is difficult to understand about defining $x$ and $b$ like this. – Matt Dec 18 '17 at 00:04
  • There is another confusing part so $a \equiv 2 mod 5$, but if we choose a = 2 then $a^2+1$ is prime. – ElChavoDelOcho Dec 18 '17 at 00:20
  • @ElChavoDelOcho You need to show that you have infinitely many choices of $a$; just take $a\equiv2\pmod{5}$ with $a>2$. – egreg Dec 18 '17 at 00:22
  • That part is showing to you that there are infinite choices of even $a$ such that $a^2 + 1$ is not prime. – Matt Dec 18 '17 at 00:27
  • @MattS , if you don’t understand what’s confusing about it, then you might not be the best person to answer the question... I find I can help people much better when I “get” where they’re stuck. – G Tony Jacobs Dec 18 '17 at 00:45
  • @GTonyJacobs This is true. If I had initially thought that was the problem, I wouldn't have answered. The question as posed asked about $r$ and $s$, which was something I did think I understood. – Matt Dec 18 '17 at 00:52
  • Looking more closely at the proof, how is it clear that $x$ and $b$ are integers? Must $r$ and $s$ have the same parity? – G Tony Jacobs Dec 18 '17 at 00:54
  • I believe $r$ and $s$ must both be odd, as $a^2 +1$ is odd. – Matt Dec 18 '17 at 00:56
  • If $a^2 + 1$ is odd then $r$ and $s$ are both odd. However the proof would work just as well if $r$ and $s$ are both even i.e. $a^2 + 1$ is divisible by $8$. – fleablood Dec 18 '17 at 02:52
  • And the proof will also work if $s =1 $ and $r = a^2 + 1$. So $a^2 +1$ *could be prime. Letting $a^2 + 1 = 17$ you get $80 = 8^2 + 4^2$ and $81 = 9^2 + 0^2$ and $82 = 9^2 + 1^2$. – fleablood Dec 18 '17 at 02:54
  • Ah. I didn’t see the part where $a$ is even. – G Tony Jacobs Dec 19 '17 at 04:33
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Just follow the instructions:

  1. Pick some even $a$ such that $a^2+1$ is not a prime, for instance some $a\equiv 2\pmod{5}$.
    Fine, let us pick $a=22$;
  2. Write $a^2+1$ as $r\cdot s$ with $r\geq s >1$.
    Fine, $22^2+1 = 97\cdot 5$;
  3. Write $a^2+1$ as a difference of squares through the previous decomposition.
    Fine, $22^2+1 = (51+46)(51-46) = 51^2-46^2$.

This gives that $51^2-1$ is the sum of two squares, $22^2+46^2$, so $51^2-1,51^2$ and $51^2+1^2$ are three consecutive numbers belonging to $\square+\square$. Is it a bit clearer now?

Jack D'Aurizio
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Let $a^2 + 1$ be odd and $a^2 + 1$ not be prime. The proof glibly assumes such numbers are easy to find by pointing out any $2 + 5k$ will be such a number.

Since $a^2 + 1$ is not prime nor then $a^2 + 1 = r*s$ for some odd $r, s$ and $s \le r$.

Let $x$ be the midpoint between $s$ and $r$. (In other words $\frac {r+s}2$). This midpoint is an integer because $r$ and $s$ are both odd.

Let $b = x-s = r-s$. This means $r = x+b$ and $s= x-b$. That's not surprising as $x$ is the midpoint and $b$ is the distance each is from the midpoint.

Consider the three consecutive numbers $n = x^2 -1$, $n+1 = x^2$ and $n + 2 = ^2 + 1$.

Obviously $n+1 = x^2 + 0^2$ is the sum of two squares. And so $n+2=x^2 + 1^2$.

But $n$ is as well because:

$a^2 + b^2 = (a^2 + 1) + b^2 - 1$

$= r*s + b^2 -1= (x+b)(x-b) + b^2 - 1$

$= x^2 -b^2 + b^2 -1 = x^2 - 1 = n$

So $n$ is the sum of $a^2+ b^2$.

fleablood
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