Question Number 88207 by MAB last updated on 09/Apr/20
![what is the biggest prime p verifying: p=Σ_(k=1) ^n [(√k)] where n∈N and [x] is floor(x)](https://www.tinkutara.com/question/Q88207.png)
$${what}\:{is}\:{the}\:{biggest}\:{prime}\:{p}\:{verifying}: \\ $$$${p}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\sqrt{{k}}\right] \\ $$$${where}\:{n}\in\mathbb{N}\:\mathrm{and}\:\:\left[{x}\right]\:{is}\:{floor}\left({x}\right) \\ $$
Commented by MJS last updated on 09/Apr/20
$$\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{197} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{working}\:\mathrm{on}\:\mathrm{an}\:\mathrm{explanation} \\ $$
Answered by MJS last updated on 09/Apr/20
![r=[(√k)]; k, q ∈N 1≤k<4 ⇒ r=1 4≤k<9 ⇒ r=2 9≤k<16 ⇒ r=3 q^2 ≤k<(q+1)^2 ⇒ r=q we get q [2q+1] times look at the sums p_q =Σ_(k=1) ^((q+1)^2 −1) [(√k)] p_1 =3 now we start adding 2s ⇒ uneven p′s p_2 =13 now adding 3s ⇒ even & uneven p′s p_3 =34=2×17 now adding 4s ⇒ even p′s p_4 =70=2×5×7 now adding 5s ⇒ 5∣p p_5 =125=5^3 now adding 6s ⇒ even & uneven p′s p_6 =203=7×29 now adding 7s ⇒ 7∣p p_7 =308=2^2 ×7×11 now adding 8s ⇒ even p′s p_8 =444=2^2 ×3×37 now adding 9s ⇒ 3∣p p_9 =615=3×5×41 now adding 10s ⇒ 5∣p p_(10) =825=3×5^2 ×11 now adding 11s ⇒ 11∣p ... I don′t have the time to continue we must prove that gcd (p_q , q+1) ≠1 ∀q>6](https://www.tinkutara.com/question/Q88248.png)
$${r}=\left[\sqrt{{k}}\right];\:{k},\:{q}\:\in\mathbb{N} \\ $$$$\mathrm{1}\leqslant{k}<\mathrm{4}\:\Rightarrow\:{r}=\mathrm{1} \\ $$$$\mathrm{4}\leqslant{k}<\mathrm{9}\:\Rightarrow\:{r}=\mathrm{2} \\ $$$$\mathrm{9}\leqslant{k}<\mathrm{16}\:\Rightarrow\:{r}=\mathrm{3} \\ $$$${q}^{\mathrm{2}} \leqslant{k}<\left({q}+\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow\:{r}={q} \\ $$$$\mathrm{we}\:\mathrm{get}\:{q}\:\left[\mathrm{2}{q}+\mathrm{1}\right]\:\mathrm{times} \\ $$$$\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{sums}\:{p}_{{q}} =\underset{{k}=\mathrm{1}} {\overset{\left({q}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}} {\sum}}\left[\sqrt{{k}}\right] \\ $$$${p}_{\mathrm{1}} =\mathrm{3} \\ $$$$\:\:\mathrm{now}\:\mathrm{we}\:\mathrm{start}\:\mathrm{adding}\:\mathrm{2s}\:\Rightarrow\:\mathrm{uneven}\:{p}'\mathrm{s} \\ $$$${p}_{\mathrm{2}} =\mathrm{13} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{3s}\:\Rightarrow\:\mathrm{even}\:\&\:\mathrm{uneven}\:{p}'\mathrm{s} \\ $$$${p}_{\mathrm{3}} =\mathrm{34}=\mathrm{2}×\mathrm{17} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{4s}\:\Rightarrow\:\mathrm{even}\:{p}'\mathrm{s} \\ $$$${p}_{\mathrm{4}} =\mathrm{70}=\mathrm{2}×\mathrm{5}×\mathrm{7} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{5s}\:\Rightarrow\:\mathrm{5}\mid{p} \\ $$$${p}_{\mathrm{5}} =\mathrm{125}=\mathrm{5}^{\mathrm{3}} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{6s}\:\Rightarrow\:\mathrm{even}\:\&\:\mathrm{uneven}\:{p}'\mathrm{s} \\ $$$${p}_{\mathrm{6}} =\mathrm{203}=\mathrm{7}×\mathrm{29} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{7s}\:\Rightarrow\:\mathrm{7}\mid{p} \\ $$$${p}_{\mathrm{7}} =\mathrm{308}=\mathrm{2}^{\mathrm{2}} ×\mathrm{7}×\mathrm{11} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{8s}\:\Rightarrow\:\mathrm{even}\:{p}'\mathrm{s} \\ $$$${p}_{\mathrm{8}} =\mathrm{444}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{37} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{9s}\:\Rightarrow\:\mathrm{3}\mid{p} \\ $$$${p}_{\mathrm{9}} =\mathrm{615}=\mathrm{3}×\mathrm{5}×\mathrm{41} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{10s}\:\Rightarrow\:\mathrm{5}\mid{p} \\ $$$${p}_{\mathrm{10}} =\mathrm{825}=\mathrm{3}×\mathrm{5}^{\mathrm{2}} ×\mathrm{11} \\ $$$$\:\:\mathrm{now}\:\mathrm{adding}\:\mathrm{11s}\:\Rightarrow\:\mathrm{11}\mid{p} \\ $$$$… \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{the}\:\mathrm{time}\:\mathrm{to}\:\mathrm{continue} \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{gcd}\:\left({p}_{{q}} ,\:{q}+\mathrm{1}\right)\:\neq\mathrm{1}\:\forall{q}>\mathrm{6} \\ $$
Commented by MAB last updated on 18/May/20
$${you}\:{are}\:{in}\:{the}\:{right}\:{way},\:{continue}. \\ $$$${hint}:\:{why}\:{don}'{t}\:{you}\:{use}\:{valuation}? \\ $$