The following solution is pretty similar to the one above but a bit "dumbed down" if you will: This question is the same question as "Prove that for any positive integer m there exists an integer $n_0$ such that for all $n > n_0, \lfloor{10\sqrt{n^2+817n+m}}\rfloor$ has a constant final digit. We note that $10\sqrt{n^2+817n+m} = \sqrt{100n^2+81700n+100m}$. Since we are using the floor function, we want to bound this somehow. Note that $(10n+40849)^2 = 100n^2+81698n+(48409)^2$ $(10n+40850)^2 = 100n^2 + 81700n + (48450)^2$ $(10n+40851)^2 = 100n^2 + 81702n + (40851)^2$ From here, we split this problem into 3 cases. $100m < 40850^2, 100m = 40850^2, 100m > 48450^2$ or $m < 4085^2, m=4085^2, m>4085^2$. Case 1: $m < 4085^2$. If this is true then $100n^2 + 81700+ 100m < 100n^2 + 81700 + 40850^2$ for all $n$. For a sufficiently large $n$ we also have $100n^2 + 81700n + 100m > 100n^2 + 81698n + 40849^2 \implies 2n > 40849^2 - 100m$ Specifically, $n > \frac{40849^2-100m}{2}$. Thus, for sufficiently large $n$ we have that $ \lfloor{10\sqrt{n^2+817n+m}}\rfloor = \sqrt{100n^2+81698n+40849^2} = 10n+40849$ which will of course always end in $9$, proving the existence $n_0 = \lfloor{\frac{40849^2-100m}{2}}\rfloor$. Case 2: $m = 4085^2$ Then $10\sqrt{n^2+817n+4085^2} = \sqrt{100n^2+81700n+40850^2} = 10n+40850$ which ends in 0 for all $n \in \mathbb{N}$, so $n_0 = 0$. Case 3: $m > 4085^2$. Then $100n^2+81700n+100 m > 100n^2 + 81700n + 40850^2$ for all $n \in \mathbb{N}$. And also, for a sufficiently large n we will have $100n^2 + 81700n+ m < 100n^2 + 81702n + 40851^2 \implies 2n > 100m-40851^2$. So for a sufficiently large $n$ we will have that $ \lfloor{10\sqrt{n^2+817n+m}}\rfloor = \sqrt{100n^2+81700n+40850^2} = 10n+40850$ which will always end in 0. Thus $n_0 = \lfloor{\frac{100m-40851^2}{2}}\rfloor$ or $0$ if $m < 4085.1^2$. Thus we have proven that $n_0$ exists in all cases. (As seen on: http://artofproblemsolving.com/community/c66163h1080420_math_and_bash_3) (shameless advertising for the blog lel)