## solve equation

IMO, IMC, USAMO, TST, Putnam, post your own math challenges to the group.

### solve equation

determine all functions $f:R \rightarrow\ R$ such that :
$f(x+f(x+y))=f(x-y)+f(x)^2$, for each $x,y \in R$, hint please
dennislewis
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Try giving x and y various values.
For example, substituting x = f(0), y = -f(0) into the equation gives f(f(0) + f(0)) = f(f(0) + f(0)) + (f(f(0)))².
Hence f(f(0)) = 0.
skipjack
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### Re: solve equation

I only get f(f(0)) = 0 from that.

With x = y = 0 you can get two possibilities for f(0); eventually I was able to make one lead to contradiction leaving only f(0) = 0.

Once you have that, take y = -x to find f(2x) = f(x) - f(x)^2. If I had continuity I feel that this would give me a function for each choice of (say) f(1), but I don't have that.
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/~bill/mingw/PARI-2-6.exe

CRGreathouse
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Location: UTC -5

Thanks. I've corrected my previous post. Now for a better attempt . . .

For any real u, substituting x = u + f(u), y = -f(u) gives f(u + 2f(u)) = (f(u + f(u)))² + f(u + 2f(u)), so f(u + f(u)) = 0.
Also, substituting x = u, y = 0 gives f(u + f(u)) = (f(u))² + f(u), so f(u)(f(u) + 1) = 0, i.e. f(u) = 0 or -1.

If f(0) = -1, substituting x = 0, y = u in the equation gives f(f(u)) = 1 + f(-u), which implies f(-u) = -1 and f(f(u)) = 0, but that gives both f(-1) = -1 and f(-1) = 0 (by putting u = 1 and u = 0, respectively), a contradiction. Hence f(0) = 0.

Putting y = -x gives f(x + f(0)) = (f(x))² + f(2x), i.e. f(x) = (f(x))² + f(2x), which implies f(x) cannot be -1.
Hence f(x) = 0 for all x.
skipjack
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