Solution:-

Let us assume the two numbers be, P and Q.

As per the condition given in the question,

\[{{P}^{2}}~+\text{ }{{Q}^{2}}~=\text{ }656\] …

[equation (i)]

P – Q = 4

P = 4 + Q … [equation (ii)]

Now, substitute the equation (ii) in equation (i) we get,

\[{{\left( 4\text{ }+\text{ }Q \right)}^{2}}~+\text{ }{{Q}^{2}}~=\text{ }656\]

We know that, \[{{\left( a\text{ }+\text{ }b \right)}^{2}}~=\text{ }{{a}^{2}}~+\text{ }2ab\text{ }+\text{ }{{b}^{2}}\]

\[({{4}^{2}}_{~}+\text{ }\left( 2\text{ }\times \text{ }4\text{ }\times \text{ }Q \right)\text{ }+\text{ }{{Q}^{2}})\text{ }+\text{ }{{Q}^{2}}~=\text{ }656\]

\[16\text{ }+\text{ }8Q\text{ }+\text{ }{{Q}^{2}}~+\text{ }{{Q}^{2}}~=\text{ }656\]

\[16\text{ }+\text{ }8Q\text{ }+\text{ }2{{Q}^{2}}~=\text{ }656\]

By transposing we get,

\[2{{Q}^{2}}~+\text{ }8Q\text{ }+\text{ }16\text{ }\text{ }656\text{ }=\text{ }0\]

\[2{{Q}^{2}}~+\text{ }8Q\text{ }\text{ }640\text{ }=\text{ }0\]

Divide both side by  \[2\]  we get,

\[2{{Q}^{2}}/2\text{ }+8Q/2\text{ }\text{ }640/2\text{ }=\text{ }0/2\]

\[{{Q}^{2}}~+\text{ }4Q\text{ }\text{ }320\text{ }=\text{ }0\]

\[{{Q}^{2}}~+\text{ }20Q\text{ }\text{ }16Q\text{ }\text{ }320\text{ }=\text{ }0\]

Take out common in each terms,

$Q(Q + 20) – 16(Q + 20) = 0$

$(Q + 20) (Q – 16) = 0$

Equate both to zero,

$Q + 20 = 0$

$Q – 16 = 0$

$Q = – 20$

$Q = 16$

$Q = 16$ … [because Q can’t be negative]

So, $P = 4 + Q$

$P = 4 + 16$

$P = 20$

Therefore, the numbers are $16 and 20$.