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Question: We want to show that if $3\nmid a,b \in \mathbb{Z} \implies a^2+b^2$ is not perfect square of an integer.

Answer: We have:

  • $3\nmid a \iff a\neq3k,\ \forall k \in \mathbb{Z}$
  • $3\nmid b \iff b\neq3l,\ \forall l \in \mathbb{Z}$

From this and the devision algorithm we have:

  • $a=3q_1+r_1,$ with $1 \leq r_1 \leq2$ for some $\ q_1,r_1\in\mathbb{Z}$

  • $b=3q_2+r_2,$ with $1 \leq r_2 \leq2$ for some $\ q_2,r_2\in\mathbb{Z}$

So, there are $4$ possible cases. I will do only the first, because I don't know if I work correctly: $$a=3q_1+1,\ b=3q_2+1$$

Lets assume that $a^2+b^2$ is perfect square of a random number $n$.

Then, $a^2+b^2=(3q_1+1)^2+(3q_2+1)^2=[3(q_1+q_2)]^2+6(q_1+q_2)+2=n^2$, and if $q_1+q_2=x$ we can say that:

$$(3x)^2+6x+2=n^2 \iff (3x+1)^2+1=n^2$$

My question is can we say now that we have contradiction? And why?

Thank you.

3 Answers 3

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So you showed that $a^2+b^2$ is a multiple of $3$ plus $2$. Can this happen? What happens when you square $3k$ or $3k\pm 1$?

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    Thank you for your answer. I had the following thought: We said that $a^2+b^2=3π+2 \iff a^2+b^2=2 (\mod 3) \iff [a^2]+[b^2]=[2]$ inside $\mathbb{Z}_3$. Is this in the right way? – 2017-01-23
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    Oh, if you *do* know modular arithmetic (and you know how to work in $\Bbb Z_3$ as a ring), then the other answer to your question does it. The point is that $[m^2]$ can only be $[0]$ or $[1]$ in $\Bbb Z_3$. My answer was giving you the same result without the "mod" language. – 2017-01-23
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    So, we can consider the two cases of $[a^2]=[0],[1]$ and proove it with contradiction. But, could you explain me your anser? Why $a^2+b^2$ is not a multiple of 3 plus 2? – 2017-01-23
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    $a^2 + b^2$ *IS* a multiple of 3 plus 2. But $x^2$ is *never* a multiple of 3 plus 2. It isn't obvious or intuitive but it is easy to figure out. because $(3x + r)^2 = 3(3x^2 + 2r) + r^2$ and $r^2$ is either $0, 1$ or $4 = 3+1$. It is never $2$ or $5$. To realize that is the entire trick. – 2017-01-23
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    To put is simply. If $3 \not \mid a$ then $a^2 = 3k + 1$ (prove it). So if $3 \not \mid a, b$ then $a^2 + b^2 = 3j + 2$. But if $3|m$ then $m^2 = 9k^2 \ne 3j + 2$. and if $3 \not \mid m$ then $m^2 = 3k +1 \ne 3j + 2$. So $a^2 +b^2 = 3j +2 \ne m^2$. – 2017-01-23
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we have $$x\equiv 0,1,2\mod 3$$ and after squaring we get $$x^2\equiv 0,1\mod 3$$

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    I would surmise that the OP has not yet learned modular arithmetic. – 2017-01-23
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    Fair enough. Then x = 3k or 3k + 1 or 3k +2 after squaring we get $x^2 = 9x^2= 3(3x^2) + 0$ or $x^2 = 9x^2 + 6x + 1 = 3(x^2 + 2x) + 1$ or $x^2 = 9x^2 + 6x + 4 = 3(x^2 + 2x + 1) + 1$. So $x^2$ must equal $3M$ or $3M + 1$ but never $3M + 2$. – 2017-01-23
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You can continue. Let $3x + 1$ = $m$ so you have $m^2 + 1 = n^2$ or two consecutive integers are both squares. That seems wrong somehow. But we must prove it.

which we can

Notice there are no integers between $n$ and $n+1$. So there are no squares between $n^2$ and $n^2 + 2n + 1$. So if $n^2 < n^2 + 1 = m^2 \le n^2 + 2n + 1$ that is only possible if $n^2 + 1 = n^2 + 2n +1$ which is only possible if $n=0$. but we assumed $3\not \mid n$.

But it might be easier to do it directly.

$a = 3k + r_1; b = 3j + r_2;$ and $m = 3l + q$ and

$a^2 + b^2 = m^2 \implies 9(k^2 + j)^2 + 6(kr_1 + jr_2) + r_1^2 + r_2^2 = 9l^2 + 6l + q^2$ or

$r_1^2 + r_2^3 \equiv q^2 \mod 3$

or $\{1,2\}^2 + \{1,2\}^2 \equiv \{0,1,2\}^2 \mod 3$

so $\{1,4\} + \{1,4\} \equiv \{0,1,4\} \mod 3$

so $\{2,5,8\} \equiv \{0,1 , 4\} \mod 3$

so $2 \equiv \{0,1\} \mod 3$ which is impossible.

And that would be a lot less of a headache if you you used $-1, 0 , + 1$ rather than $0,1 ,2$.

The $a = 3k \pm 1$ and $b = 3j \pm 1$ and $m = 3l \pm 1$ or $ 3l$.

So $a^2 + b^2 = 9(k^2 + j^2) \pm 6(j+k) + 1 + 1 \equiv 2 \mod 3$ while

$m^2 = 9l^2 \pm 6l +\{0, 1\}\equiv 0, 1 \mod 3$. A contradiction.

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    Thank you for your answer. I can understand better the middle answer. (I think that you have a typo in the exponents) – 2017-01-23
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    You the exponents should all be 2. So if I had a 3... it was a typo. – 2017-01-23
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    The middle part is definitely easier and I'm sure it is the intended solution. But I believe strongly in letting a student figure it out and follow ones own path. And your path *was* valid and *did* get to $(3x + 1)^2 + 1 = n^2$ which *is* a contradiction. Just not an immediate and obvious one. – 2017-01-23
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    See something else: We said that exists an $n$ such that $a^2+b^2=n^2$ and after my claim in the question this happens if and only if $3|n^2-2 \iff [n]\cdot [n]=[2]$ in $\mathbb{Z_3}$ and this is a contradiction because we there not exists such an $n$. Is this right? (I apologize for my English) – 2017-01-23
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    Yes. Actually using modular arithmetic the argument is the absolute easiest. All integers are congruent so something mod 3 so if $a^2 + b^2 = n^2$ then $[a]^2 + [b]^2 = [n]^2$ There are only three elements in $\mathbb Z_3$ [0], [1] and [2]. $[0]^2 = 0; [1]^2 = 1; [2]^2 = 1$. So $[i]^2 = 2$ is impossible. if $[i] \ne [0]$ then [i]= 1 or [i]=2 but in either case $[i]^2 = 1$ so $[a]^2 + [b]^2 = 2$. And $[m]^2 = 2$ is impossible. That's the easiest. – 2017-01-23