[proofplan] We prove the identity directly in $\mathbb{Z}$ by expanding the two squares on the right-hand side. The cross terms cancel, leaving exactly the product $(a^2+b^2)(c^2+d^2)$. Substituting the hypotheses $m=a^2+b^2$ and $n=c^2+d^2$ then gives the asserted representation of $mn$ as a sum of two integer squares. [/proofplan] [step:Verify that the proposed summands are integers] Define \begin{align*} r &:= ac - bd, \\ s &:= ad + bc. \end{align*} Since $a,b,c,d \in \mathbb{Z}$ and $\mathbb{Z}$ is closed under addition, subtraction, and multiplication, we have $r,s \in \mathbb{Z}$. Thus $r^2+s^2$ is a [sum of two squares](/theorems/1725) in $\mathbb{Z}$. [/step] [step:Expand the proposed sum of squares and cancel the cross terms] Using distributivity and commutativity of multiplication in $\mathbb{Z}$, we compute \begin{align*} (ac-bd)^2 + (ad+bc)^2 &= \left(a^2c^2 - 2abcd + b^2d^2\right) + \left(a^2d^2 + 2abcd + b^2c^2\right) \\ &= a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 \\ &= a^2(c^2+d^2) + b^2(c^2+d^2) \\ &= (a^2+b^2)(c^2+d^2). \end{align*} [/step] [step:Substitute the hypotheses to obtain the representation of $mn$] By the hypotheses, \begin{align*} m = a^2+b^2, \qquad n = c^2+d^2. \end{align*} Substituting these equalities into the identity established above gives \begin{align*} (ac-bd)^2 + (ad+bc)^2 = (a^2+b^2)(c^2+d^2) = mn. \end{align*} Since $ac-bd$ and $ad+bc$ are integers, this proves that $mn$ is a sum of two squares in $\mathbb{Z}$, with the explicit representation \begin{align*} mn = (ac-bd)^2 + (ad+bc)^2. \end{align*} [/step]