If ## a ## is an odd integer, then ## a^{2}\equiv 1\pmod {8} ##?

[Math100]

Homework Statement

Prove the assertion below:
If $a$ is an odd integer, then $a^{2}\equiv 1\pmod {8}$.

Relevant Equations

None.

Proof:

Suppose $a$ is an odd integer.
Then $a=2k+1$ for some $k\in\mathbb{Z}$.
Note that $a^{2}=(2k+1)^{2}=4k^{2}+4k+1=4k(k+1)+1$.
Since $k(k+1)$ is the product of two consecutive integers,
it follows that $k(k+1)$ must be even.
This means $k(k+1)=2m$ for some $m\in\mathbb{Z}$.
Thus $a^{2}=4(2m)+1=8m+1\implies a^{2}\equiv 1\pmod {8}$.
Therefore, if $a$ is an odd integer, then $a^{2}\equiv 1\pmod {8}$.

 

[Delta2]

I think this is mostly correct, not sure if we should take separately the case $a=1$ (which is obvious that it holds) but for $a=1$ the $k$ in the proof is zero so I am not sure if $k(k+1)=0(0+1)=0$ can be said to be even in this case.

 

[fresh_42]

I think this is mostly correct, not sure if we should take separately the case $a=1$ (which is obvious that it holds) but for $a=1$ the $k$ in the proof is zero so I am not sure if $k(k+1)=0(0+1)=0$ can be said to be even in this case.

Why not? $0=2\cdot 0$ is even. It has to be since $2\mathbb{Z}$ is a subgroup of $\mathbb{Z}.$