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  1. Find all functions $f$, defined on the real numbers and taking real values, which satisfy the equation $f(x)f(y) = f(x + y) + xy$ for all real numbers $x$ and $y$.

Thanks in advance for any contributions.

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    $\begingroup$ What are your thoughts/attempts on the problem? $\endgroup$ – Martin R Mar 25 '15 at 10:33
  • $\begingroup$ Well I worked out f(0)=1, and f(-1)f(1)=0, but then I hit a wall. $\endgroup$ – MadChickenMan Mar 25 '15 at 10:35
  • $\begingroup$ That is basically the method I've been using, but I can't see anything I can prove by induction?! $\endgroup$ – MadChickenMan Mar 25 '15 at 10:43
  • $\begingroup$ $f(1)f(-1) = 0$ means $f(1) = 0$ or $f(-1) = 0$... $\endgroup$ – kennytm Mar 25 '15 at 10:47
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Setting $y = 0$ gives $f(x)f(0) = f(x)$. Since $f$ cannot be identically zero, it follows that $f(0) = 1$.

Setting $x = 1, y= -1$ then gives $f(1)f(-1) = 0$, therefore $f(1) = 0$ or $f(-1) = 0$ must hold.

In the first case, setting $x = u-1, y = 1$ gives $0 = f(u) + u-1$ $\Longleftrightarrow $ $\boxed{f(u) = 1 - u} \,$.

I'll leave it to you to verify that this is really a solution, and to investigate the second case.

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  • $\begingroup$ Thank you, indeed this does lead to a solution, but should you not have to substitute x back in, or is this arbitrary. $\endgroup$ – MadChickenMan Mar 25 '15 at 12:18
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    $\begingroup$ @MadChickenMan: You can name the variable whatever you want: $f(x) = 1 - x$, $f(y) = 1 - y$, $f(u) = 1 - u$ is all the same. Strictly speaking the function is $f: \mathbb R \to \mathbb R, x \to 1-x$ and it does not matter if you use $x$ or something else. $\endgroup$ – Martin R Mar 25 '15 at 12:19
  • $\begingroup$ Ok thanks. Pardon my inexperience, but what does f:R>R mean, I've seen it used a lot. $\endgroup$ – MadChickenMan Mar 25 '15 at 12:24
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    $\begingroup$ @MadChickenMan: It means that $f$ is a mapping from $\mathbb R$ into $\mathbb R$, i.e. the domain is $\mathbb R$ and the values are also in $\mathbb R$. $\endgroup$ – Martin R Mar 25 '15 at 12:28

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