Question

The function f satisfies the equation f(x)+f(y)=f(x+y)-xy-1 for every pair x, y among all real numbers. If f(1)=1, find all integers n (if any) such that f(n)=n

Answer 1

i would think this way for a start
f(x+y)=f(x)+f(y)+xy+1
f(1)=1
f(2)=4
So far i'm thinking of a sequence or something appearing just a feeling though

Answer 2

How do you suppose f(2)=4?

Answer 3

we are given f(1)
i just made x=y in that case, but again I'm just trying to see anything useful

Answer 4

hmm
let's see f(n)=n ==> f(n)+f(n)=f(2n)-n^2-1
2f(n)=f(2n)-n^2-1
f(2n)=n^2+2n+1=(n+1)^2

Answer 5

maybe we can use f(1)=1 first to get
f(x)+f(1-x)=-x(1-x)
then use that f(n)=n where n is an integer to have
x+(1-x)=-x(1-x) then solve this equation and see if we get x as an integer...

Answer 6

interesting @myininaya
I haven't thought of using both, but i'm not in your league guys hehe

Answer 7

x^2-x-1=0 <== this has no integer solutions

Answer 8

okay we have this-> \(f(1)=1\)
now
\(f(1)+f(0)=f(1+0)-1 \times 0 -1\)
from here we get-> \(f(0)=-1\)

now
\(f(-1)+f(1)=f(-1+1)-1 \times 1 -1\)
\(f(-1)+1=-1 -1 -1\)
\(f(-1)=-4\)

now for any general number say \(a\)

\(f(a-1) + f(a+1)=f(2a) - (a-1)(a+1) -1\) ----> equation p

now i do this-
lets say \(n\) is the number for which \(f(n)=n\)
so
equation 1 equation 2
\(\color{blueviolet}{f(n)+f(1)=f(n+1)-n-1}\) \(\color{orange}{f(n)+f(-1)=f(n-1)+n-1}\)

now add equation 2 and equation 1
we get this-

\(2f(n) +f(1)+f(-1)=f(n-1)+f(n+1)-2\)

now putting \(f(n)=n\) and using equation p we can write->

\(2n+1-4=f(2n) -(n-1)(n+1)-1\)

now @xapproachesinfinity showed that \(f(2n)=(n+1)^2\) putting that in our equation we get-

\(2n+1-4=(n+1)^2-n^2+1-1\)
simplifying we get-
\(2n-3=1+2n\)
:)
\(-3=1\)
so no n possible o-0 except 1?

Answer 9

wow so the equation arrived by @myininaya is giving the same hint
no solution except the given one f(1)=1

Answer 10

I think @imqwerty is getting it. Keep going.