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If the statement is true, then prove it, otherwise provide a counter example.

If $x,y \in \mathbb{Z}$ such that $x = y^2$, then $x\equiv y \pmod 2$

Could someone just check to see if my proof is correct or that i made a mistake somewhere. Thank you.

Proof: Counter Example. There exists $x,y \in \mathbb{Z}$ such that $x=y^2$, then $x \not \equiv y \pmod 2$. Let $x$ equal $3$, then $y^2$ equals $9$. Because you cannot divide $9$ by $2$, the statement $x \equiv y \pmod2$ is false. $3 \not \equiv 3 \pmod2$.

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user3365751
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  • Why do you think that $9\not \equiv 3\pmod 2$? -- In fact, if you take $x=3$ and leave $y$ unspecified while assuming $x=y^2$, why do you think that $y^2=9$? – Hagen von Eitzen Nov 06 '17 at 17:56
  • You've misunderstood the definition of the relation $x \equiv y \pmod n$. You're asserting that $n \not\mid y$, which is true but not relevant. You need to decide if $n \mid (x-y)$, as Siong Thye Goh notes. – Matthew Leingang Nov 06 '17 at 18:02
  • Congruence modulo two is just a question of parity (evenness or oddness). Two things are $\equiv\pmod2$ if and only if they have the same parity: both are even or both are odd. Given that outlook I think you can answer the question in a handful of words. – Lubin Nov 06 '17 at 18:19

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If $x=9$ and $y=3$, we actually have $9 \equiv 3 \pmod2$ because we can divide $9\color{red}{-3}$ by $2$.

Try to check that if you square a number, you do not change the parity. If you square an odd number, you get an odd number. If you square an even number, you get an even number.

Siong Thye Goh
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We essentially need $y(y-1)$ to be divisible by $2$

which is obvious as the product of two consecutive integers is always even.

More generally, we can prove $$y^n\equiv y\pmod2$$

lab bhattacharjee
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