# $2+ f(x)f(y)=f(x)+f(y) +f(xy)$, if $f(2)=5$ find $f(5)$

This question has been asked before, however I am interested in seeing why my approach to finding a solution does not work.

$$2+ f(x)f(y)=f(x)+f(y) +f(xy)$$, if $$f(2)=5$$ find $$f(5)$$

What I have done is the following:

Plugging in 0 gives us a quadratic which can be solved to give the values of $$f(0)=1, 2$$

Now if I take $$x=5, y=0$$ using $$f(0)=1$$ gives us $$1=1$$ Which is not useful.

Using $$f(0)=2$$ we get $$f(5)=2$$.

This is not the correct answer, which is given as $$26$$.

Why is this an incorrect method to go about solving the question?

• Taking $x=5, y=0$ with $f(0)=1$ gives $2+f(5)\times 1=f(5)+1+1\implies 2=2$. No problem. – lulu Apr 13 at 10:25
• Why not include the link to the prior version of the question? – lulu Apr 13 at 10:27
• You just included a link to your own question. The earlier question assumes that $f$ is continuous (actually, polynomial) which, as far as I can tell, is an absolutely essential assumption. – lulu Apr 13 at 10:40
• "Using $f(0)=2$ we get $f(5)=2$." I'm going to guess that your mistake is in there somewhere, but you've not really shown us what you did (nor in the earlier question). – Teepeemm Apr 23 at 23:09

What appears to be incorrect in your method is that you seem to be ruling out the possibility that $$f(0)=1$$ because it leads to a result that is "not useful." In a logical argument, you can only rule something out if it leads to a contradiction. The conclusion $$1=1$$, while indeed not useful, is nonetheless true. So $$f(0)=1$$ remains a viable option.
Indeed, as Kavi Rama Murthy's answer shows, it's the possibility $$f(0)=2$$ that leads to a contradiction: Letting $$x=0$$ and $$y=2$$ in the equation $$2+f(x)f(y)=f(x)+f(y)+f(xy)$$ gives
$$f(0)=2\implies 2+2f(2)=2+f(2)+2\implies f(2)=2\not=5$$
However, knowing now that $$f(0)$$ must equal $$1$$ is, as you put it, not useful. All it says is that $$2+f(y)=1+f(y)+1$$. But neither is the conclusion $$f(1)=2$$, which you can derive from $$2+f(1)f(2)=f(1)+f(2)+f(1\cdot2)$$, because all it'll tell you is that $$2+2f(y)=2+f(y)+f(y)$$.
In point of fact, nothing you do here will prove useful, because the problem, as stated, has dropped an assumption from the question linked to in the OP, namely that the function $$f$$ be a polynomial. Without an additional assumption on the nature of the function, $$f(5)$$ can have literally any value.
$$f(0)=2$$ does not lead to valid solution to the equation. If $$f(0)=2$$ put $$x=0$$ to get $$2+2f(y)=4+f(y)$$ or $$f(y)=2$$ for all $$y$$. This is in conflict with the given condition $$f(2)=5$$ so we cannot take $$f(0)=2$$.