Prove that $\forall n \in \mathbb{N}_+,\exists p,q \in \mathbb{Z}$ such that $|p^2+2q^2-n|<\sqrt[4]{9n}$
Solution. 1) Assertion is true $\forall n\in[1,108]$ proof: Let $p\in\{0,1,2,3,4,5\}$ If $n\in[3p^2,3p^2+2p+1)$ : $\min(n-(2p^2+p^2),(2p^2+(p+1)^2)-n)\le p<\sqrt[4]{27p^2}<\sqrt[4]{9n}$ If $n\in[3p^2+2p+1,3p^2+4p+2)$ : $\min(n-(2p^2+(p+1)^2),(2(p+1)^2+p^2)-n)\le p<\sqrt[4]{27p^2}<\sqrt[4]{9n}$ If $n\in[3p^2+4p+2,3(p+1)^2]$ : $\min(n-(2(p+1)^2+p^2),(2(p+1)^2+(p+1)^2)-n)\le p<\sqrt[4]{27p^2}<\sqrt[4]{9n}$ Q.E.D.
2) Assertion is true $\forall n\in [108,289]$ Proof: Assertion is true $\forall n\in\{108,118\}$ $\min(n-(2.2^2+10^2),(2.3^2+10^2)-n)\le 5<\sqrt[4]{9\times 108}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{118,129\}$ $\min(n-(2.3^2+10^2),(2.8^2+1^2)-n)\le 5<\sqrt[4]{9\times 118}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{129,139\}$ $\min(n-(2.8^2+1^2),(2.3^2+11^2)-n)\le 5<\sqrt[4]{9\times 129}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{139,150\}$ $\min(n-(2.3^2+11^2),(2.5^2+10^2)-n)\le 5<\sqrt[4]{9\times 139}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{150,162\}$ $\min(n-(2.5^2+10^2),(2.3^2+12^2)-n)\le 6<\sqrt[4]{9\times 150}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{162,172\}$ $\min(n-(2.3^2+12^2),(2.6^2+10^2)-n)\le 5<\sqrt[4]{9\times 162}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{172,179\}$ $\min(n-(2.6^2+10^2),(2.7^2+9^2)-n)\le 3<\sqrt[4]{9\times 172}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{179,187\}$ $\min(n-(2.7^2+9^2),(2.9^2+5^2)-n)\le 4<\sqrt[4]{9\times 179}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{187,198\}$ $\min(n-(2.9^2+5^2),(2.1^2+14^2)-n)\le 5<\sqrt[4]{9\times 187}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{198,209\}$ $\min(n-(2.1^2+14^2),(2.10^2+3^2)-n)\le 5<\sqrt[4]{9\times 198}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{209,219\}$ $\min(n-(2.10^2+3^2),(2.5^2+13^2)-n)\le 5<\sqrt[4]{9\times 209}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{219,228\}$ $\min(n-(2.5^2+13^2),(2.4^2+14^2)-n)\le 5<\sqrt[4]{9\times 219}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{228,241\}$ $\min(n-(2.4^2+14^2),(2.6^2+13^2)-n)\le 6<\sqrt[4]{9\times 228}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{241,251\}$ $\min(n-(2.6^2+13^2),(2.11^2+3^2)-n)\le 5<\sqrt[4]{9\times 241}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{251,264\}$ $\min(n-(2.11^2+3^2),(2.2^2+16^2)-n)\le 6<\sqrt[4]{9\times 251}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{264,275\}$ $\min(n-(2.2^2+16^2),(2.5^2+15^2)-n)\le 6<\sqrt[4]{9\times 264}\le\sqrt[4]{9n}$
Assertion is true $\forall n\in\{275,289\}$ $\min(n-(2.5^2+15^2),(2.12^2+15^2)-n)\le 7<\sqrt[4]{9\times 275}\le\sqrt[4]{9n}$
3) Assertion is true $\forall n\ge 290$ Proof: Let $p=\left\lfloor\sqrt{\frac{n-1}2}\right\rfloor$ so that $n\in\{2p^2+1,2(p+1)^2\}$ So $n\in[2p^2+k^2,2p^2+(k+1)^2)$ for some $k\in\{1,2,...,\left\lfloor\sqrt{4p+3}\right\rfloor\}$
And so $\min(n-2p^2-k^2,2p^2+(k+1)^2-n)\le k$ $\le \left\lfloor\sqrt{4p+3}\right\rfloor$
And so it remains to prove that : $\left\lfloor\sqrt{4p+3}\right\rfloor<\sqrt[4]{9n}$ where $p=\left\lfloor\sqrt{\frac{n-1}2}\right\rfloor$
Enough to prove $\sqrt{4\sqrt{\frac{n-1}2}+3}<\sqrt[4]{9n}$, which is easily proved true $\forall n\ge 290$ Q.E.D.
I hope someone will post a nicer solution !