Let $ N = 100a+10b+c$, where $ a, b, c\in\mathcal{D}_{10}=\{0, 1,\ldots, 9\}$. Since $ 11\mid N$, then $ b = a+c$ and $ N = 110a+11c$, yielding $ N/11 = 10a+c$. In fact, we need $ 10a+c = a^{2}+(a+c)^{2}+c^{2}= 2(a^{2}+ac+c^{2})$. Hence $ c = 2d$, for a suitable $ d\in\{0, 1,\ldots, 4\}$, and consequently $ 5a = a^{2}+2ad+(4d-1)\cdot d$. This is impossible if $ d\ge 2$, since then $ \mbox{RHS}\ge a^{2}+2a+3 > 5a$. So necessarily $ d = 0$ and $ a^{2}= 5a$, i.e. $ a = 0$ or $ a = 5$, i.e. $ N = 0$ or $ N = 550$. As $ 0 = 0^{2}$ and $ 50 = 5^{2}+5^{2}$, we're done.

s.tringali wrote: Let $ N = 100a + 10b + c$, where $ a, b, c\in\mathcal{D}_{10} = \{0, 1,\ldots, 9\}$. Since $ 11\mid N$, then $ b = a + c$ and $ N = 110a + 11c$, yielding $ N/11 = 10a + c$. In fact, we need $ 10a + c = a^{2} + (a + c)^{2} + c^{2} = 2(a^{2} + ac + c^{2})$. Hence $ c = 2d$, for a suitable $ d\in\{0, 1,\ldots, 4\}$, and consequently $ 5a = a^{2} + 2ad + (4d - 1)\cdot d$. This is impossible if $ d\ge 2$, since then $ \mbox{RHS}\ge a^{2} + 2a + 3 > 5a$. So necessarily $ d = 0$ and $ a^{2} = 5a$, i.e. $ a = 0$ or $ a = 5$, i.e. $ N = 0$ or $ N = 550$. As $ 0 = 0^{2}$ and $ 50 = 5^{2} + 5^{2}$, we're done. How about 803?

s.tringali wrote: Since $ 11\mid N$, then $ b = a + c$ The error is here. As Peter has demonstrated, it is also possible that $ b + 11 = a + c$.