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In general $(x+y)^2 \neq (x^2+y^2)$. Is this still correct in the case of mod $2$? Can anyone share some insight?

tonix
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  • The generalized answers given below are great, but it may help to consider a particular example to solidify why the negated equivalence doesn't hold in $\mod 2$. Consider when $x=y=1$. Then $$(x+y)^2=(1+1)^2=(0)^2=0,$$ and $$(x^2+y^2)=(1^2+1^2)=(1+1)=0.$$ Thus, here is a simple counterexample to your claim that $(x+y)^2\neq(x^2+y^2)$ for the integers $\mod 2$. –  Mar 02 '15 at 08:07
  • And then $(0+0)^2 = 0^2 + 0^2$ is a counterexample for the general case ... ? – user133281 Mar 02 '15 at 08:07
  • @user133281 If your comment is directed to me, then please indicate this. I'm not sure what you're trying to say. –  Mar 02 '15 at 08:09
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    @therapist I think user133281 is trying to point out that you are proving the statement with the wrong quantifier. You need to prove $\forall$ but your example proves $\exists$. I.e when we say $(x+y)^2\neq x^2+y^2$ we mean there $\exists x,y$ s.t. $(x+y)^2\neq x^2+y^2$. So to disprove this you must show $\forall x,y; (x+y)^2=x^2+y^2$ – DRF Mar 02 '15 at 08:19
  • @DRF Maybe I'm reading the question wrong then. It seems like OP is claiming the negated equality holds for all integers mod $2$. My counterexanple was to show this is not true. –  Mar 02 '15 at 08:50
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    @therapist That way of reading the question makes no sense, since for no field is it true that $\forall x,y;;(x+y)^2\neq x^2+y^2$, as user 133281 points out. Thus the only sensible reading of the question is "Is it true for fields of characteristic 2 that $\exists x,y;;(x+y)^2\neq x^2+y^2$?" It sounds strange due to the phrasing which uses negation. – DRF Mar 02 '15 at 09:16

4 Answers4

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Not so because $x^2 + y^2 = x^2 + 2xy + y^2 = (x+y)^2 \pmod 2$ as $ 2xy = 0 \pmod 2$

DeepSea
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That statement is false in the case of mod 2. In fact the case of $(x+y)^p=x^p+y^p$ is true mod p for all prime values of $p$.

ASKASK
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By the binomial theorem

$$(x+y)^n=x^n+y^n+\sum_{k=1}^{n-1}{n\choose k}x^k y^{n-k}$$

Now if $n$ is prime, $n$ divides ${n\choose k}$ if $0<k<n$ because the factor $n$ appears in the numerator of

$${n\choose k} = \frac{n\times (n-1) \times \dots \times (n-k+1)}{k!}$$

but not the denominator (as $k<n$ and $n$ is prime) (Source).

Therefore, for $n$ prime, ${n\choose k}$ is a multiple of $n$ and under the operation $\pmod n$ the rightmost sum in the first formula vanishes, meaning

$$(x+y)^n \equiv x^n+y^n \pmod n$$

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If you want to read about this, I suggest you to take the book Abstract Algebra in the Chapter 13 (Field Theory, third edition) and check about characteristic of a field. Here prove that ASKASK says.

Jonathan G.
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