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Solve the functional equation
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
+
2
f
(
x
)
f
(
y
)
∀
x
,
y
∈
R
{\displaystyle f(x+y)=f(x)+f(y)+2{\sqrt {f(x)f(y)}}\quad \forall {x,y\in \mathbb {R} }}
If this equation is true for any y , then it is true for the particular case y = 0. Thus:
f
(
x
)
=
f
(
x
)
+
f
(
0
)
+
2
f
(
x
)
f
(
0
)
{\displaystyle f(x)=f(x)+f(0)+2{\sqrt {f(x)f(0)}}}
⟹
−
f
(
0
)
=
2
f
(
x
)
f
(
0
)
{\displaystyle \implies -f(0)=2{\sqrt {f(x)f(0)}}}
⟹
f
(
0
)
2
=
4
f
(
x
)
f
(
0
)
⟹
f
(
0
)
=
0
∨
4
f
(
x
)
=
f
(
0
)
∀
x
∈
R
{\displaystyle \implies f(0)^{2}=4f(x)f(0)\implies f(0)=0\lor 4f(x)=f(0)\quad \forall {x\in \mathbb {R} }}
Now, if
4
f
(
x
)
=
f
(
0
)
{\displaystyle 4f(x)=f(0)}
x , then it is also true for x = 0, which means that 4 f(0) = f(0), thus f(0) = 0 and
f
(
x
)
=
0
{\displaystyle f(x)=0}
Let a >0 be a real number with the value f(1). We now prove that any solution must be of the form
f
(
x
)
=
a
x
2
{\displaystyle f(x)=ax^{2}}
f
(
0
)
=
0
{\displaystyle f(0)=0}
f
(
1
)
=
a
{\displaystyle f(1)=a}
f
(
x
)
=
a
x
2
{\displaystyle f(x)=ax^{2}}
n integers. Then,
f
(
n
+
1
)
=
f
(
n
)
+
f
(
1
)
+
2
f
(
n
)
f
(
1
)
=
a
n
2
+
a
+
2
a
n
2
a
=
a
(
n
2
+
1
+
2
n
)
=
a
(
n
+
1
)
2
{\displaystyle f(n+1)=f(n)+f(1)+2{\sqrt {f(n)f(1)}}=an^{2}+a+2{\sqrt {an^{2}a}}=a(n^{2}+1+2n)=a(n+1)^{2}}
Now we simplify the original equation into
f
(
x
+
y
)
=
(
f
(
x
)
+
f
(
y
)
)
2
{\displaystyle f(x+y)=\left({\sqrt {f(x)}}+{\sqrt {f(y)}}\right)^{2}}
a
=
f
(
1
)
=
f
(
p
−
1
p
+
1
p
)
=
(
f
(
p
−
1
p
)
+
f
(
1
p
)
)
2
{\displaystyle a=f(1)=f\left({\frac {p-1}{p}}+{\frac {1}{p}}\right)=\left({\sqrt {f\left({\frac {p-1}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}}
=
(
(
f
(
p
−
2
p
)
+
f
(
1
p
)
)
2
+
f
(
1
p
)
)
2
{\displaystyle =\left({\sqrt {\left({\sqrt {f\left({\frac {p-2}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}}
=
(
f
(
p
−
2
p
)
+
f
(
1
p
)
+
f
(
1
p
)
)
2
{\displaystyle =\left({\sqrt {f\left({\frac {p-2}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}}
=
⋯
{\displaystyle =\cdots }
=
(
p
f
(
1
p
)
)
2
{\displaystyle =\left(p{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}}
f
(
1
p
)
=
a
1
p
2
{\displaystyle f\left({\frac {1}{p}}\right)=a\ {\frac {1}{p^{2}}}}
