Question 1209940
<pre>

{{{f(x-y) = sqrt(f(x*y)+4*x+4*y-8)}}}

{{{f(x-y) = sqrt(f(x*y)+4*x+4*y-8)}}}

If we can find a case of x and y where {{{f(x-y)=f(x*y)=f(1)}}}, then we could
solve for f(1).  That would require  

{{{x-y=x*y=1}}}

{{{y=1/x}}}, {{{x-1/x=1}}}

{{{x-1/x=1}}}
{{{x^2-1=x}}}
{{{x^2-x-1=0}}}
{{{x=(1 +- sqrt(1+4))/2}}} <--we can only use the + sign
{{{x=(1+sqrt(5))/2}}}

Incidentally, that, or its reciprocal, is the golden ratio, famous in 
historical architecture. Let's call it {{{G=(1+sqrt(5))/2}}}.

{{{G=(1+sqrt(5))/2}}} and {{{G-1/G=1}}}

{{{f(G-1/G) = sqrt(f(G*expr(1/G))+4*G+4*expr(1/G))}}}

{{{f(1) = sqrt(f(1)+4*(G+1/G))}}}

Since {{{G-1/G=1}}}, {{{1/G=G-1}}}

{{{f(1) = sqrt(f(1)+4*(G+G-1))}}}

{{{f(1) = sqrt(f(1)+4*(2G-1))}}}

{{{f(1) = sqrt(f(1)+8G-4)}}}

{{{(f(1)^"")^2 = f(1)+8G-4}}}

{{{(f(1)^"")^2 - f(1)-(8G-4)=0}}}

Solve for f(1)

{{{f(1)=(1 +- sqrt(1+4(8G-4)))/2}}}  <--- we can only use +

{{{f(1)=(1 + sqrt(1+32G-16))/2}}}

{{{f(1)=(1 + sqrt(32G-15))/2}}}

Substituting {{{G=(1+sqrt(5))/2}}}, and doing some algebra,
it simplifies to

{{{f(1)=(1+sqrt(1+16sqrt(5)))/2}}}

{{{f(1)=3.532205783}}} approximately.

Edwin</pre>