39yes it is the first one only.
now i edited it.
This functional equation is nice and has a super short and elegant solution:
Find all functions f: N --> N such that
xf(y) + yf(x) = (x + y)f(x2 + y2)
for all x, y and in N, where N = {0, 1, 2, 3, ...}.
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13 Answers
62arrey na...
I was asking if the questoin is
f(x2 + y2)
or f(2x+2y)
I assume it is the first one?
1Putting y=0 : f(x2) = f(0) which shows f is constant
so f(x) is any whole number
66from f(x2)=f(0), how do you conclude that f(x) is also a constant? It just proves that when x = 0, 1, 4, 9, 25 etc i.e. perfect squares then f(x)=f(0). but what about other values os x like 2, 3, 5, 6 , ...?
106Find all functions f: N --> N such that
xf(y) + yf(x) = (x + y)f(x2 + y2)
for all x, y and in N, where N = {0, 1, 2, 3, ...}.
xf(0) = xf(x2)
i.e. f(x2)=f(0)
√xf(0) = √xf(x)
==> f(x) = f(0)
and as f:N→N so f(0) must be whole no.
62he has replace x2 by x
I strongly think that this is wrong
What is correct is only uptil f(x2) = f(0)
so he has essentially proved that this is true for all perfect squares..
39the solution is as follows:
If f(x) is not constant then we can find x, y suxh that f(x) - f(y) has least positive integer value.
But then
f(y) < f(x2+y2) = (xf(y) + yf(x))/(x+y) < f(x)
And we get the contradiction.
So f(x) must be constant function which in fact satisfies the equation.
1my bad for not figuring cases for x=2,3,5..
and i was coming with the solution but bhargav already did that :)