An Olympiad Problem from Poland 🇵🇱

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❤️ f(x + y) - f(x - y) = f(x)f(y) Poland MO

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PLAYLISTS 🎵 :

[krishnanadityan]

After all the arguments, we prove that f(x)=0

[sngmn]

Replace y with -y and with given eqn we will get a system:
{ f(x+y) - f(x-y) = f(x)f(y)
{ f(x-y) - f(x+y) = f(x)f(-y)
Sum these eqns:
0 = f(x)(f(y) + f(-y))
Suppose f(x) = 0 then its done.
Suppose f(y) + f(-y) = 0 then f(y) - odd function, stop what? But we already prove that f(x) is even(at 4:11) ))) So f(x) and even and odd function. What function it can be? Right, f(x) = 0.

[ChristopherEvenstar]

I was able to follow the steps. It went great! I love watching the functional equation videos. I feel like my brain gets more flexible. Thanks!

[grchauvet]

I hoped there was a non-trivial solution for some other field - e.g. one of characteristic 2, but the proof is even quicker: For any x, we have 0 = f(0)-f(0) = f(x+x)-f(x-x) = f(x)^2, and hence f(x) = 0.

[cicik]

- substitute 0, 0 : f(0) - f(0) = f²(0) => f(0) = 0
- substitute 0, n:f(n) -f(-n) = f(0)f(n) =0 so f(n) = -f(n), so f is symmetrical

- substitute x = a+b, y = a-b:
f(a) * f(b) = f(2x) - f(2y) = f(b+a)(b-a) (because f is symmetrical) = f(2y) - f(2x) so
2f(2x) = 2f(2y) or f is constant, but we figured out above that f = 0
- answer: f(anything) = 0

[majora]

I finished up from step 4 in a slightly different way.

• -f(2x) = f(x) * f(-x)
• -f(2x) = f(-x) * f(-x) from step 2
• -f(2x) = [f(-x)]^2
• -f(2x) = f(-2x) from step 3


Previously we deduced that f(x) is even, and we just now deduced f(x) is odd too. The only function that's both even and odd at the same time is f(x) = 0.

[Cow.cool.]

i found that f(x) was zero but kept trying because i thought there was another solution. Its like a math rickroll

[Sparkles.]

Could you please say, how one would know that the answer satisfies the condition...Thank you😊

[davidsousaRJ]

If we substitute y = 0 and keep x we get f(x) = 0 as a solution since the beginning. But then we have to keep trying other substitutions until we verify no other solutions are possible.

[orchestra]

Beautiful👍 I did it in a bit different way though. I figured out that f(0)=0, but missed out on the even/odd function argument
However, I noticed that if we put y=∆x and ∆x->0, then in the limit the LHS gives 2∆x*f'(x) and the RHS gives f(x)*f'(0)*∆x, because This all of course rests on the assumption that this limit exists, and f is differentiable at x.
Then, we have 2 f'(x) = f(x)*f'(0). The solution of this ODE is f(x) = f(0)*exp(f'(0)*x/2). But because f(0)=0, then f(x) = 0 for all x.

[paulortega]

So close. If it had only been f(x+y) + f(x-y) = f(x)*f(y).

[kengo]

f(0)=0
(f(x+h)-f(x-h))/(2h) = f(x)/2 * f(h)/h
h -> 0
f’(x) = f(x)/2 * (f(h) - 0)/(h - 0) = f(x)/2 * (f(0 + h) - f(0))/(h - 0) = f(x)/2 * f’(0)
But when x = 0
f’(0) = f(0)/2 * f’(0) = 0 because f(0) = 0
f’(x) = 0
==> f(x) = 0

[anotherelvis]

f(x)*f(0)=f(x+0)-f(x-0)=0 => f(0)=0
f(0+y)-f(0-y)=f(y)*f(0)=0 => f(y)=f(-y)
f(x+x)-f(x-x)=f(x)*f(x) => f(2x)+0=f(x)^2
f(x-x)-f(x-(-x))=f(x)*f(-x) => -f(2x)-0=f(x)*(-f(x))= f(x)^2
So
f(2x)=f(x)^2-f(2x) => f(2x)=0

[tontonbeber]

f(x+y)-f(x-y) = f(x)f(y)
(a) y=0 => f(x)-f(x) = 0 = f(x)f(0)
=> f(x)=0 or f(y)=0 the second in inside the first, so f(0)=0
(b) x=0 => f(y)-f(-y) = f(0)f(y) = 0 => f(y)=f(-y)
(c) y=x => f(2x)-f(0) = f(2x)=f(x)²
(d) y=-x => f(0)-f(2x) = -f(2x) = f(x)f(-x) = (b) f(x)²
(c, d) => f(2x)=-f(2x) => f(2x)=0 and so f(x)=0 forall x

[leonidfedyakov]

This video should be shortened to 2 minutes. Too many words.

[andypandy]

What a useless function that is always zero.. :D

[alextang]

f(x+y)-f(x-y)=f(x)*f(y)
Put x=x, y=0
f(x)-f(x)=f(x)*f(0)
f(0)*f(x)=0
Therefore f(0)=0 or f(x)=0
In short f(x)=0 answer. 😋😋😋😋😋😋

[DutchMathematician]

The way I solved it.
In such questions it is often beneficial to insert x=0, y=0 (or both), as well as x=y and/or x=-y.

I proceeded as follows.
Fill in x=0 and y=0. This gives:
f(0)-f(0)=0=f(0)*f(0)
Hence f(0)=0.
Substituting x=y, we get (for arbitrary x):
f(2*x)-f(0)=f(x)*f(x)
or:
f(2*x)=f(x)*f(x)
This means that f(2*x) is the product of two identical real numbers, hence f(2*x) is non-negative. Since x was arbitrary, we can conclude that f(2*x) is non-negative for all x and hence the same holds for f(x).
Now substitute y=-x in the general equation. We get:
f(0)-f(2*x)=f(x)*f(-x)
or (since f(0)=0):
-f(2*x)=f(x)*f(-x)
Since we've concluded that f(x)≥0 for every x, this only can mean that f(x)=0 for every x.

[dominiquelarchey-wendling]

f(x+y)-f(x-y) = f(x)f(y)

x, y := 0 gives f(0)-f(0) = f(0)f(0), hence f(0)² = 0. Thus f(0) = 0
x := 0 gives f(y)-f(-y) = f(0)f(y) = 0. Hence f(y) = f(-y) for any y.

y := -y gives f(x-y)-f(x+y) = f(x)f(-y) = f(x)f(y) for any x, y

Thus we have for any x, y we have

f(x+y)-f(x-y) = f(x)f(y)
f(x-y)-f(x+y) = f(x)f(y)

adding both gives

0 = 2f(x)f(y) hence f(x)f(y) = 0 for any x, y

Hence with y := x, we get f(x)² = 0, thus f(x) = 0 for any x