0

Prove that:

If $x$ is an integer in the form $x=n^2+2$ for some integer $n$, then $x$ can also be expressed in the form of $4m+2$ or $4m+3$ for some integer $m$.

May I know what is the possible approach for this proof?

Aiden Chow
  • 2,849
  • 10
  • 32
Emily W
  • 65
  • 4
  • 1
    You mean $4m+2$ (if $n$ and $x$ are even) or $4m+3$ (if $n$ and $x$ are odd)? – J. W. Tanner Sep 27 '20 at 01:03
  • It should be $4m+2$ or $4m+3$, not and. The two are incompatible. The simplest approach is to look at the squares $\bmod 4$. There are not many of them. – Ross Millikan Sep 27 '20 at 01:04
  • Hint: $n$ can be written as either $n=2k$ or $n=2k+1$ – Andrei Sep 27 '20 at 01:07
  • Exploit parity: $\bmod 4!:\ \color{#0a0}{\rm even}^2\equiv 0,\ \rm \color{#c00}{odd}^2\equiv 1, $ i.e. $ x\equiv \color{#0a0}{0,2},\color{#c00}{1,3}\Rightarrow x^2\equiv \color{#0a0}0,\color{#c00}1 $ so $ x^2!+!2\equiv \color{#0a0}2,\color{#c00}3 $. More generally we have $ {\rm odd}^2\equiv 1\pmod{!8}, $ which is frequently useful. – Bill Dubuque Sep 27 '20 at 01:43

1 Answers1

0

Note that for any two $a, b\in \Bbb N$, we can find a $ q\in\Bbb N_0, r\in\{0,\ldots,b-1\}$ such that $a=qb+r$.

As a result of this important theorem, for any integer $n$ we have that $n=2k$ or $n=2k+1$, and also that $n=3k$ or $n=3k+1$ or $n=3k+2$, and so on and so forth. $k$ is, of course, an integer in every case.

Number theory proofs very often, if not almost always, rely on splitting the integers into groups in this fashion and proving that each case works. In this case, if we use that $n=2k$ or $n=2k+1$, we can see that $$(2k)^2+2=4k^2+2=4m+2 \text{ (where $m=k^2$)}$$ and $$(2k+1)^2+2=4k^2+4k+3=4m+3\text{ (where $m=k^2+k$)}$$

Rhys Hughes
  • 12,842