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For $1<c<c_0$, for some constant $c_0$ the circle method will yield an asymptotic of size $$E(A)\sim \mathfrak{S}N^{4/c-1.}$$ where $\mathfrak{S}$ is a constant. In fact, when $c$ is not too large, one can even obtain an asymptotic for the additive energy of the Piatetski-Shapiro primes (see Balog and Friedlander). Letting $S_{A}(\theta)=\mathbb{E}_{n\leq N}1_{A}(n)e(n\theta)$ we may write $$E(A)=N^{3}\int_{0}^{1}|S_{A}(\theta)|^{4}d\theta,$$ or alternatively $$E(A)=\frac{1}{N}\int_{0}^{1}\left|\sum_{n\leq N^{1/c}}e(\lfloor n^{c}\rfloor\theta)\right|^{4}d\theta.$$

Remark: I am not sure how large $c_0$ can be, but the asymptotic definitely breaks down for $c=2$.

For $1<c<c_0$, for some constant $c_0$ the circle method will yield an asymptotic of size $$E(A)\sim \mathfrak{S}N^{4/c-1.}$$ where $\mathfrak{S}$ is a constant. In fact, when $c$ is not too large, one can even obtain an asymptotic for the additive energy of the Piatetski-Shapiro primes (see Balog and Friedlander). Letting $S_{A}(\theta)=\mathbb{E}_{n\leq N}1_{A}(n)e(n\theta)$ we may write $$E(A)=N^{3}\int_{0}^{1}|S_{A}(\theta)|^{4}d\theta.$$ The sum $S_A(\theta)$ can be handled by noting that $[(n+1)^{1/c}]-[n^{1/c}]$ is the indicator function for the Piatetski-Shapiro sequence, and then using the truncated Fourier series for the sawtooth function.

For $c$ not too much greater than $1$, the circle method will yield an asymptotic of size $$E(A)\sim \mathfrak{S}N^{4/c-1.}$$ where $\mathfrak{S}$ is a constant. In fact, for small values of $c$ you can even obtain an asymptotic for the additive energy of the Piatetski-Shapiro primes (see Balog and Friedlander). Letting $S_{A}(\theta)=\mathbb{E}_{n\leq N}1_{A}(n)e(n\theta)$ we may write $$E(A)=N^{3}\int_{0}^{1}|S_{A}(\theta)|^{4}d\theta,$$ or alternatively $$E(A)=\frac{1}{N}\int_{0}^{1}\left|\sum_{n\leq N^{1/c}}e(\lfloor n^{c}\rfloor\theta)\right|^{4}d\theta.$$

Remark: I am not sure for what the value of $c$ the asymptotic breaks down, but it certainly does not hold with $c=2$.

For $1<c<c_0$, for some constant $c_0$ the circle method will yield an asymptotic of size $$E(A)\sim \mathfrak{S}N^{4/c-1.}$$ where $\mathfrak{S}$ is a constant. In fact, when $c$ is not too large, one can even obtain an asymptotic for the additive energy of the Piatetski-Shapiro primes (see Balog and Friedlander). Letting $S_{A}(\theta)=\mathbb{E}_{n\leq N}1_{A}(n)e(n\theta)$ we may write $$E(A)=N^{3}\int_{0}^{1}|S_{A}(\theta)|^{4}d\theta,$$ or alternatively $$E(A)=\frac{1}{N}\int_{0}^{1}\left|\sum_{n\leq N^{1/c}}e(\lfloor n^{c}\rfloor\theta)\right|^{4}d\theta.$$

Remark: I am not sure how large $c_0$ can be, but the asymptotic definitely breaks down for $c=2$.

For small values of $c>1$, the circle method will yield an asymptotic of size $$E(A)\sim \mathfrak{S}N^{4/c-1.}$$ where $\mathfrak{S}$ is a constant. In fact, for small values of $c$ you can even obtain an asymptotic for the additive energy of the Piatetski-Shapiro primes (see Balog and Friedlander). Letting $S_{A}(\theta)=\mathbb{E}_{n\leq N}1_{A}(n)e(n\theta)$ we may write $$E(A)=N^{3}\int_{0}^{1}|S_{A}(\theta)|^{4}d\theta,$$ or alternatively $$E(A)=\frac{1}{N}\int_{0}^{1}\left|\sum_{n\leq N^{1/c}}e(\lfloor n^{c}\rfloor\theta)\right|^{4}d\theta.$$

Remark: I am not sure for what the value of $c$ the asymptotic breaks down, but it certainly does not hold with $c=2$.

For $c$ not too much greater than $1$, the circle method will yield an asymptotic of size $$E(A)\sim \mathfrak{S}N^{4/c-1.}$$ where $\mathfrak{S}$ is a constant. In fact, for small values of $c$ you can even obtain an asymptotic for the additive energy of the Piatetski-Shapiro primes (see Balog and Friedlander). Letting $S_{A}(\theta)=\mathbb{E}_{n\leq N}1_{A}(n)e(n\theta)$ we may write $$E(A)=N^{3}\int_{0}^{1}|S_{A}(\theta)|^{4}d\theta,$$ or alternatively $$E(A)=\frac{1}{N}\int_{0}^{1}\left|\sum_{n\leq N^{1/c}}e(\lfloor n^{c}\rfloor\theta)\right|^{4}d\theta.$$

Remark: I am not sure for what the value of $c$ the asymptotic breaks down, but it certainly does not hold with $c=2$.