If x is an odd number, how to prove $x^2=8y+1$?

Question

I tried to let $x=2k+1$ but it ended up proving $k^2+k=2y$ for some $k$ and $y$. What is the correct solution?

See also: If $n$ is an odd natural number, then $8$ divides $n^{2}-1$ and $n\text{ odd}\implies n^2=8k+1$ for some $k\in \mathbb{Z}$ — Martin Sleziak, Dec 30 '18 at 08:49

score 2 · Answer 1 · answered Feb 07 '18 at 19:04

2

HINT

Assume that $x=2k+1$ then $$x^2 = 4k(k+1)+1$$

answered Feb 07 '18 at 19:04

gt6989b

54,422

This is exactly what I did but I don’t see the ration of this and 8y+1 – Randyy Feb 07 '18 at 19:05
It's because one of $k$ or $k+1$ is even. – fred goodman Feb 07 '18 at 21:00

score 1 · Answer 2 · answered Feb 07 '18 at 19:05

1

Hint: use that $\,x^2-1=(x-1)(x+1)$, and one of two consecutive evens is divisible by $\,4\,$.

answered Feb 07 '18 at 19:05

dxiv

76,497

score 1 · Answer 3 · answered Feb 07 '18 at 19:05

1

If $x$ is odd, then it is congruent to $1$, $3$, $5$ or $7$, modulo $8$. Since $$ 1^2\equiv 1,\quad 3^2\equiv 1,\quad 5^2\equiv 1,\quad 7^2\equiv1\pmod{8} $$ you're done.

For the “hard” approach, consider that $k(k+1)$ is even.

answered Feb 07 '18 at 19:05

egreg

238,574

I know this statement is correct and x-1 is always a multiple of 8 but can you write this on the test? Feel like it’s not a strict proof – Randyy Feb 07 '18 at 19:06
@MingzhouZhu Sorry, but I can’t know what you’re allowed to use in your test. This is certainly a full proof nonetheless. – egreg Feb 07 '18 at 20:30

score 0 · Accepted Answer · answered Feb 07 '18 at 19:25

0

Yet another approach, an odd number $x$ is congruent to 1 or 3 mod 4, so try the two cases $x = 4 k + 1$ and $x = 4 k + 3$ and compute $x^2$.

answered Feb 07 '18 at 19:25

fred goodman

4,253

If x is an odd number, how to prove $x^2=8y+1$?

4 Answers4