We claim that the only answer to this equation is $f(x)\equiv x^2+\frac{1}{4}$, which could be easily verified to be an answer. Let $P(x,y)$ denote the given equation for $x,y\in\mathbb{Q}.$ First, consider $P(x,0)\, ; \, f\left(\frac{1}{2}\right)+f(x)=4f(0)f(x)+\frac{1}{2}\,\forall\, x\in\mathbb{Q}$ if $f(0)\neq\frac{1}{4}$, then $f(x)$ must be constant which is clearly a contradiction, thus $f(0)=\frac{1}{4}$ which also implies that $f\left(\frac{1}{2}\right)=\frac{1}{2}$. Next, consider $P\left(x,\frac{1}{2}\right)\, ; \, f\left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=4f\left(\frac{1}{2}\right)f(x)+\frac{1}{2}=2f(x)+\frac{1}{2}\,\forall\, x\in\mathbb{Q}$ Solving the recurrence equation, we can see that for every $x\in\mathbb{Q},$ there exists two rational numbers $b_x$ and $c_x$ such that $f(t)=t^2+b_xt+c_x$ for any rational number $t$ such that $t-x$ is half of an integer number. (In particular, $f(x)=x^2+\frac{1}{4}$ for all $x$ which is half of an integer number.) Finally, fix a rational number $q$, then for any integer $r$ such that $qr\in\mathbb{Z},$ we have $P(q,r)\, ; \, f\left(2qr+\frac{1}{2}\right)+f(q-r)=4f(q)f(r)+\frac{1}{2}$ which could be rewritten as $$\left(2qr+\frac{1}{2}\right)^2+\frac{1}{4}+(q-r)^2+b_q(q-r)+c_q=(4r^2+1)(q^2+b_qq+c_q)+\frac{1}{2}$$or $$(4q^2+1)r^2-b_qr+\left(q^2+b_qq+c_q+\frac{1}{2}\right)=4(q^2+b_qq+c_q)r^2+\left(q^2+b_qq+c_q+\frac{1}{2}\right)$$Since this equation is true for infinitely many integers $r$, we must have $4q^2+1=4(q^2+b_qq+c_q)$, thus $f(q)=q^2+b_qq+c_q=q^2+\frac{1}{4}$ as desired.

Here is easier solution: Let $P(x,y)$ be the given equation.If $f(0)\neq \frac{1}{4}$ then $P(x,0)$ $\implies$ $f$ is constant which has no solution.$P(x,0)$ also gives that $f(\frac{1}{2})=\frac{1}{2}$. Take a function $g(x)=4f(\frac{x}{2})$ then original equation becomes $g(4xy+1)+g(2x-2y)=g(2x)g(2y)+1$ Then replace $(x,y)$ by $(\frac{x}{2},\frac{y}{2})$ and let this equation be $Q(x,y)$: $g(xy+1)+g(x-y)=g(x)g(y)+2$ So $g(0)=1$,$g(1)=2$. $Q(x,y),Q(y,x)$ yields that $g$ is even.$Q(x,1)$ gives that $g(x+1)+g(x-1)=2g(x)+2$ $(1)$ which means $g(n)=n^2+1$ for all integers by easy induction. $\textbf{Claim}$:$g(x+n)+g(x-n)=2g(x)+2n^2$ for all natural $n$ and rational $x$. Proof:We are going to prove that by induction on $n$.Base case is clear by $(1)$. Again by $(1)$ $g(x+n+1)+g(x-n-1)=2g(x+n)+2-g(x+n-1)+2g(x-n)+2-g(x-n+1)=2(2g(x)+2n^2)+4-2g(x)-2(n-1)^2=2g(x)+2(n+1)^2$. $Q(x,n) + Q(x,-n) $ gives that $g(xn+1)+g(xn-1)+g(x+n)+g(x-n)=2g(x)(n^2+1)+4=2g(nx)+2+2g(x)+2n^2$ $\implies$ $g(nx)=n^2g(x)-(n^2-1)$ So take $x=\frac{m}{n}$ then $g(\frac{m}{n})=(\frac{m}{n})^2+1$ which means $g(x)=x^2+1$ which means that $f(x)=x^2+\frac{1}{4}$ which clearly satisfies equation.

The answer is $\boxed{f(x)=x^2+\frac{1}{4}}$. It is clear that this works; now we prove that it is the only solution. Let $P(x,y)$ be the given assertion; and we derive $f(0)=\frac{1}{4}$, $f \left(\frac{1}{2}\right)=\frac{1}{2}$ as #2, #3 did. $P \left(x, \frac{1}{2} \right)$ gives $f\left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=2f(x)+\frac{1}{2}$. Since $f(x)=x^2+\frac{1}{4}$ holds for $x=0 , \frac{1}{2}$, and $f(x)=x^2+\frac{1}{4}$ is a solution of the above equation; so $f\left(\frac{n}{2}\right)=\left(\frac{n}{2}\right)^2+\frac{1}{4}$ for every $n \in \mathbb{Z}$. $P\left(x, \frac{n}{2}\right)$ gives $f\left(nx+\frac{1}{2}\right)+f\left(x-\frac{n}{2}\right)=\left(n^2+1\right)f(x)+\frac{1}{2}$. Plugging in $x=\frac{p}{q}$ and $n=q$ gives: $f\left(p+\frac{1}{2}\right)+f\left(\frac{p}{q}-\frac{q}{2}\right)=\left(q^2+1\right)f\left(\frac{p}{q}\right)+\frac{1}{2}$. Plugging in $x=\frac{p}{q}-\frac{q}{2}$ and $n=-q$ gives $f\left(-p+\frac{q^2}{2}-\frac{1}{2}\right)+f\left(\frac{p}{q}\right)=\left(q^2+1\right)f\left(\frac{p}{q}-\frac{q}{2}\right)+\frac{1}{2}$. Now note that $f\left(p+\frac{1}{2}\right)$ and $f\left(-p+\frac{q^2}{2}-\frac{1}{2}\right)$ are both on the form $f\left(\frac{\text{integer}}{2}\right)$; so it is equal to $f(\cdot)=(\cdot)^2+\frac{1}{4}$. Plugging in these values and solving the system of equation gives $f\left(\frac{p}{q}\right)=\left(\frac{p}{q}\right)^2+\frac{1}{4}$ for all $\frac{p}{q} \in \mathbb{Q}$, as desired.

lminsl wrote: The answer is $\boxed{f(x)=x^2+\frac{1}{4}}$. It is clear that this works; now we prove that it is the only solution. Let $P(x,y)$ be the given assertion; and we derive $f(0)=\frac{1}{4}$, $f \left(\frac{1}{2}\right)=\frac{1}{2}$ as #2, #3 did. $P \left(x, \frac{1}{2} \right)$ gives $f\left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=2f(x)+\frac{1}{2}$. After this ,take $x\to x+\frac{1}{2},$ and $x\to x-\frac{1}{2}$ we have $f(x-1)+f(x+1)=2f(x)+2,$ and with induction we get easily $f(x+n)=(n+1)f(x)-nf(x-1)+n(n+1)$ and $f(x-n)=(1-n)f(x)+nf(x-1)-n(1-n).$ Adding these two equation we get $f(x+n)+f(x-n)=2f(x)+2n^2,$ for all $x\in\mathbb{Q},n\in\mathbb{Z}.$ And from $P(\frac{m}{n},n)+P(\frac{m}{n},-n)$ we can easily get $f(\frac{m}{n})=(\frac{m}{n})^2+\frac{1}{4},$ and we are done. Note: we can find easily$f$ is even function.

$\textbf{Harder Version}$:Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that $$f(2xy + \frac{1}{2}) + f(x-y) = 4f(x)f(y) + \frac{1}{2}$$for all $x,y \in \mathbb{R}$. Solution:As I did for rationals again take $g(x)=4f(\frac{x}{2})$.And equation becomes $g(xy+1)+g(x-y)=g(x)g(y)+2$. $g(0)=0$,$g(1)=2$.so $g$ isn't constant.One can easily get that $g$ is even.Let the equation be $Q(x,y)$. $Q(x,y)-Q(x,-y)$ $\implies$ $g(x+y)-g(x-y)=g(1+xy)-g(1-xy)$.Let $x=\frac{m+n}{2}$ and $y=\frac{m-n}{2}$ for some real numbers $m,n$.We find that: $g(m)-g(n)=g(1+\frac{m^2-n^2}{4})-g(1-\frac{m^2-n^2}{4})$. Defining $h(x)=g(1+\frac{x}{4})-g(1-\frac{x}{4})$ then $g(m)-g(n)=h(m^2-n^2)$. Letting $n=0$ we deduce that $g(m)=h(m^2)+g(0)=h(m^2)+1$.Whence, $h(1)=1$. Moreover, $h(a^2-b^2)+h(b^2-c^2)=g(a)-g(b)+g(b)-g(c)=g(a)-g(c)=h(a^2-c^2)$.So $h$ is additive.So let's substitute to $Q(x,y)$ in terms of $h$: $g(x)g(y)=(h(x^2)+1)(h(y^2)+1)=g(1+xy)+g(x-y)-2=h(1+2xy+x^2y^2)+1+h(x^2-2xy+y^2)+1-2=h(1)+2h(xy)+h(x^2y^2)+h(x^2)+h(y^2)-2h(xy)=1+h(x^2y^2)+h(x^2)+h(y^2)$ $\implies$ $h(x^2y^2)=h(x^2)h(y^2)$ so $h$ is multiplicative in positive reals since $h$ is odd(because of additivity) $h$ is multiplicative in all reals.Thus $h(x)=x$ which means $g(x)=x^2+1$ so $f(x)=x^2+\frac{1}{4}$.

falantrng wrote: lminsl wrote: The answer is $\boxed{f(x)=x^2+\frac{1}{4}}$. It is clear that this works; now we prove that it is the only solution. Let $P(x,y)$ be the given assertion; and we derive $f(0)=\frac{1}{4}$, $f \left(\frac{1}{2}\right)=\frac{1}{2}$ as #2, #3 did. $P \left(x, \frac{1}{2} \right)$ gives $f\left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=2f(x)+\frac{1}{2}$. After this ,take $x\to x+\frac{1}{2},$ and $x\to x-\frac{1}{2}$ we have $f(x-1)+f(x+1)=2f(x)+2,$ and with induction we get easily $f(x+n)=(n+1)f(x)-nf(x-1)+n(n+1)$ and $f(x-n)=(1-n)f(x)+nf(x-1)-n(1-n).$ Adding these two equation we get $f(x+n)+f(x-n)=2f(x)+n^2,$ for all $x\in\mathbb{Q},n\in\mathbb{Z}.$ And from $P(\frac{m}{n},n)+P(\frac{m}{n},-n)$ we can easily get $f(\frac{m}{n})=(\frac{m}{n})^2+\frac{1}{4},$ and we are done. Note: we can find easily$f$ is even function. i guess some brackets are missing in the third line ...

Denote the assertion as $P(x,y)$. $P(x,0)$ gives $f\left(\frac{1}{2}\right)+f(x)=4f(x)f(0)+\frac{1}{2}$, so $\left(f\left(\frac{1}{2}\right),f(0)\right)=\left(\frac{1}{2},\frac{1}{4}\right)$, or $f(x)$ is constant. The latter gives $2f=4f^2+\frac{1}{2}$, which has no rational roots, so we must have the former. Now, $P\left(x,\frac{1}{2}\right)$ gives $f\left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=2f(x)+\frac{1}{2}$. So, if we induct upwards on a residue class mod $\frac{1}{2}$, we see that $f(x)=x^2+bx+c$ for some $b,c$ if $x$ is fixed $\pmod{\frac{1}{2}}$. Now, suppose we fix $x,y,xy$ mod $\frac{1}{2}$, and plug them into $P$. For suitable constants $b_i,c_i$ for $1\le i\le 4$, we get $$\left(2xy+\frac{1}{2}\right)^2+b_1\left(2xy+\frac{1}{2}\right)+c_1+(x-y)^2+b_2(x-y)+c_2=4(x^2+b_3x+c_3)(y^2+b_4y+c_4)+\frac{1}{2}$$By fixing one variable over multiple values and varying the other, we easily see that corresponding coefficients must be equal. In particular, the coefficient of $x^2y$ is $4b_4$ on the RHS and $0$ on the LHS, so $b_i=0\forall i$. Furthermore, the coefficient of $x^2$ is $1$ on the LHS and $4c_4$ on the RHS, so $c_i=\frac{1}{4}\forall i$. Thus, we must have $f(x)\equiv x^2+\frac{1}{4}$, and this can be easily verified to work.

The only function that works is $f(x)=x^2+\frac 14$. Clearly it works. Now I claim only this function works. LHS is not symmetric but RHS is, so $f(x-y)=f(y-x)$ so $f$ is even. It's clear that $f$ is nonconstant: suppose $f(x)\ne f(y)$. $P(0,x),P(0,y)$ yields $(4f(0)-1)m=f(\frac 12)-\frac 12$ having two solutions, $f(x),f(y)$, so $f(0)=\frac{1}{4}, f(\frac{1}{2})=\frac 12$ $P(\frac{1}{2},y)$ gives $f(y+\frac 12)+f(y-\frac 12)=2f(y)+\frac 12$ so $f(\frac{n}{2})=\frac{n^2+1}{4}\forall n\in\mathbb{Z}$ Suppose $xy,x\in\mathbb{Z}$ (1), then $x^2-xy\in\mathbb{Z}$ Then $P(x,y)$ and $P(-x,y)$ yields $4x^2y^2+2xy+f(x-y)=(4x^2+1)f(y)$ $4x^2(x-y)^2+2x(x-y)+f(y)=(4x^2+1)f(x-y)$ Details omitted: solving the equations yield $f(y)=y^2+\frac 14$ Note if $y=\frac{p}{q}$ then picking $x=q$ would suffice.

face you, i have! .

Let $P(x,y)$ denote the given assertion. Note that we must have $f(0)=\tfrac{1}{4}$ otherwise $P(x,0)$ suggests $f$ is constant, clearly false. Moreover (by $P(x,0)$) we have $f(\tfrac{1}{2})= \tfrac{1}{2}.$ $P(x,\tfrac{1}{2})$ implies $f(x+\tfrac{1}{2})+f(x-\tfrac{1}{2})=2f(x)+\tfrac{1}{2}.$ It follows that $f(\tfrac{x}{2})=(\tfrac{x}{2})^2+\tfrac{ax}{2}+b$ for all integer $x.$ Now take an integer $k$ such that $\tfrac{p}{q}\cdot k$ is an integer. Then $P(\tfrac{p}{q},k)$ (ofcourse we have to solve it) would force $f(\tfrac{p}{q})=(\tfrac{p}{q})^2+\tfrac{1}{4}$ we can check this works.

Let $P(x,y)$ be the assertion. Let $f(0) = c$. $P(x,0) , P(0,x) : f(x) = f(-x)$ $P(0,0) : f(\frac{1}{2}) + c = 4c^2 + \frac{1}{2} \implies f(\frac{1}{2}) = c(4c-1) + \frac{1}{2}$ $P(x,0) : c(4c-1) + \frac{1}{2} + f(x) = 4cf(x) + \frac{1}{2} \implies c = \frac{1}{4}$ or $f(x) = c$ which isn't possible so $c = \frac{1}{4}$ so $f(0) = \frac{1}{4}$ and $f(\frac{1}{2}) = \frac{1}{2}$. $P(x,\frac{1}{2}) : f(x + \frac{1}{2}) + f(x - \frac{1}{2}) = 2f(x) + \frac{1}{2}$ $P(x,\frac{1}{2}) + P(x - \frac{1}{2},\frac{1}{2}) + P(x + \frac{1}{2},\frac{1}{2}) : f(x-1) + f(x+1) = 2f(x) + 2$ Now Let $g(x) = 4f(x)$ and Let $g(2xy+\frac{1}{2}) + g(x-y) = g(x)g(y) + 2$ be the new assertion called $Q(x,y)$. we Also have that $g(x+\frac{1}{2}) + g(x-\frac{1}{2}) = 2g(x) + 2$ and $g(x+1) + g(x-1) = 2g(x) + 8$ and we show their assertion by $T(x)$ and $K(x)$. $T(\frac{1}{2}) : g(1) + 1 = 6 \implies g(1) = 5$ Now we have $g(0) = 1,g(1) = 5$ so by induction on $K(x)$ we get that for every Natural number $n$ we have $g(n) = 4n^2 + 1$. Also Note that $K(x+n-1) + K(x+n-2) + ... + K(x-n+1) : g(x+n) + g(x-n) = 2g(x) + 8n^2$ for every Natural number $n$. $Q(\frac{x}{2},n) + Q(\frac{x}{2},-n) : g(xn+\frac{1}{2}) + g(-xn+\frac{1}{2}) + g(\frac{x}{2} + n) + g(\frac{x}{2} - n) = g(x)(g(n) + g(-n)) + 4$ so $g(xn+\frac{1}{2}) + g(xn-\frac{1}{2}) + 2g(\frac{x}{2}) + 8n^2 = g(\frac{x}{2})(8n^2 + 2) + 4$ so $g(xn+\frac{1}{2}) + g(xn-\frac{1}{2}) = 8n^2g(\frac{x}{2}) - 8n^2 + 4$ so by $T(xn)$ we have $g(xn+\frac{1}{2}) + g(xn-\frac{1}{2}) = 2g(xn) + 2 = 8n^2g(\frac{x}{2}) - 8n^2 + 4$ so $g(xn) = 4n^2g(\frac{x}{2}) - 4n^2 + 1$. Let this be $R(x,n)$ so now by $R(2x,n) , R(x,2n) : g(2xn) = 4n^2g(x) - 4n^2 + 2 = 16n^2g(\frac{x}{2}) - 16n^2 + 1$ so $4n^2g(x) + 12n^2 = 16n^2g(\frac{x}{2}) \implies g(x) + 3 = 4g(\frac{x}{2})$ now putting this result in $R(x,n)$ we would get $g(xn) = n^2(g(x) + 3) - 4n^2 + 1 \implies n^2g(x) - n^2 + 1$ Now let $x = \frac{m}{n}$ for Natural $m,n$ and we would get $g(m) = n^2g(\frac{m}{n}) - n^2 + 1$ so $4m^2 + 1 + n^2 - 1 = n^2g(\frac{m}{n}) \implies 4\frac{m^2}{n^2} + 1 = g(\frac{m}{n})$ so for every $x \in \mathbb{Q}$ we have $g(x) = 4x^2 + 1$ so $f(x) = x^2 + \frac{1}{4}$ which clearly works.