If $x = a^2 + b^2 + c^2 + d^2$, show that
$$ 2x = (a + b)^2 + (a− b)^2 + (d− c)^2 + (c + d)^2 $$
Given $1984 = 40^2 + 16^2 + 8^2 + 8^2$, convert
(i) $3968 = 2 \times 1984$ into sum of four squares.
(ii) $992 = \frac{1984}{2}$ into sum of four squares. Hint: Use part (i).
We proceed as follows
$$ \begin{align} (a + b)^2 + (a− b)^2 + (d− c)^2 + (c + d)^2 & = a^2 + b^2 +\cancel{2ab} + a^2 +b^2 -\cancel{2ab} +d^2 +c^2 -\cancel{2cd} + c^2 + d^2 + \cancel{2cd} \\ \\ & = 2(a^2 + b^2 + c^2 + d^2) \\ \\ & = 2x \end{align}$$
(i) We have $1984 = 40^2 + 16^2 + 8^2 + 8^2$, and so
$$ \begin{align} 3968 & = 2 \times 1984 \\ \\ & = 2 \times (40^2 + 16^2 + 8^2 + 8^2) \\ \\ & = (40+16)^2 + (40-16)^2 + (8-8)^2 + (8+8)^2 \\ \\ & = 56^2 + 24^2 + 16^2 \end{align} $$
(ii) We proceed as follows
$$ \begin{align} 992 & = \frac{1984}{2} \\ \\ & = \frac{2(40^2 + 16^2 + 8^2 + 8^2)}{2^2} \\ \\ & = \frac{56^2 + 24^2 + 16^2}{2^2} \\ \\ & = 28^2 + 12^2 + 8^2 \end{align} $$