Product of Integers of form 4n + 1

Theorem

Let $m, n \in \Z$ such that both $m$ and $n$ are of the form $4 k + 1$ where $k \in \Z$.

Then $m n$ is also of the form $4 k + 1$.

Let $m = 4 k_1 + 1, n = 4 k_2 + 1$.

Then:

$\ds m n$

$=$

$\ds \paren {4 k_1 + 1} \paren {4 k_2 + 1}$

$\ds $

$=$

$\ds 4 k_1 \cdot 4 k_2 + 4 k_1 + 4 k_2 + 1$

$\ds $

$=$

$\ds \paren {4 k_1 \cdot 4 k_2 + 4 k_1 + 4 k_2} + 1$

$\ds $

$=$

$\ds 4 \paren {k_1 \cdot 4 k_2 + k_1 + k_2} + 1$

$\ds $

$=$

$\ds 4 \paren {4 k_1 k_2 + k_1 + k_2} + 1$

$\ds $

$=$

$\ds 4 k + 1$

where $k = 4 k_1 k_2 + k_1 + k_2 \in \Z$

$\blacksquare$

1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $8$