/bump $ $ any help?

N1 ?

I claim that $\boxed{a_i=\tfrac{7^{2^i}-1}{6}}$ works just fine Rather than simply prove my claim I will show you my motivation: zschess wrote: $ 6a_i + 1 $ is a prime power. This is a rather strong statement, and it would be hard to show unless it is a power of a specific prime. So we decide to choose a sequence of the form $a_i=\tfrac{p^{e_i}-1}{6}$ where $e_i$ is dependent on $i$. zschess wrote: $ a_{i + 1} $ is divisible by $ a_{i} $. If follows then that $e_i\mid e_{i+1}$. zschess wrote: $ a_i $ is divisible by $ 2^{i + 2} $ but not $ 2^{i + 3} $. This reminds of the LTE Lemma. Since $i+2=v_2(a_i)=v_2(p-1)+v_2(p+1)+v_2(e_i)-1-v_2(6)$ we are motivated to make $e_i=2^i$, and then choose $p$ that satisfies $v_2(p-1)+v_2(p+1)=4$ zschess wrote: $ a_i $ is not divisible by $ 3 $. Again by LTE we want $v_3(p-1)=1$ zschess wrote: $ a_i $ can be written as the sum of the two perfect squares. This is perhaps the hard one. By Fermat's Theorem it should be fine if there is no prime that is $3\text{ (mod }4)$ that divides $a_i$. Let $q$ be such a prime. Then $v_2(q-1)=1$, so we need only have $q\nmid p^2-1$ or $q=3$. We now need only find a prime $p$ such that $v_2(p-1)+v_2(p+1)=4$, $v_3(p-1)=1$, and $q\nmid p^2-1$ or $q=3$. The lowest prime we can reasonably check, $p=7$, suffices $\Box$

rafayaashary1 wrote: I claim that $a_i=\tfrac{7^{2^i}-1}{6}$ works just fine or (13^(2^(i+1)-1)/6 and lagrange will work

remark arcoding to dirichlet the progression have infinitely prime 6k+1 and rest is adijust the exponent of 2

lebathanh wrote: rafayaashary1 wrote: I claim that $a_i=\tfrac{7^{2^i}-1}{6}$ works just fine or (13^(2^(i+1)-1)/6 and lagrange will work What is $Lagrange$?

(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2

rafayaashary1 wrote: I claim that $\boxed{a_i=\tfrac{7^{2^i}-1}{6}}$ works just fine Rather than simply prove my claim I will show you my motivation: zschess wrote: $ 6a_i + 1 $ is a prime power. This is a rather strong statement, and it would be hard to show unless it is a power of a specific prime. So we decide to choose a sequence of the form $a_i=\tfrac{p^{e_i}-1}{6}$ where $e_i$ is dependent on $i$. zschess wrote: $ a_{i + 1} $ is divisible by $ a_{i} $. If follows then that $e_i\mid e_{i+1}$. zschess wrote: $ a_i $ is divisible by $ 2^{i + 2} $ but not $ 2^{i + 3} $. This reminds of the LTE Lemma. Since $i+2=v_2(a_i)=v_2(p-1)+v_2(p+1)+v_2(e_i)-1-v_2(6)$ we are motivated to make $e_i=2^i$, and then choose $p$ that satisfies $v_2(p-1)+v_2(p+1)=4$ zschess wrote: $ a_i $ is not divisible by $ 3 $. Again by LTE we want $v_3(p-1)=1$ zschess wrote: $ a_i $ can be written as the sum of the two perfect squares. This is perhaps the hard one. By Fermat's Theorem it should be fine if there is no prime that is $3\text{ (mod }4)$ that divides $a_i$. Let $q$ be such a prime. Then $v_2(q-1)=1$, so we need only have $q\nmid p^2-1$ or $q=3$. We now need only find a prime $p$ such that $v_2(p-1)+v_2(p+1)=4$, $v_3(p-1)=1$, and $q\nmid p^2-1$ or $q=3$. The lowest prime we can reasonably check, $p=7$, suffices $\Box$ but you did not prove that you have infintey sequence of positive integers

raghoodah1m wrote: rafayaashary1 wrote: I claim that $\boxed{a_i=\tfrac{7^{2^i}-1}{6}}$ works just fine but you did not prove that you have infintey sequence of positive integers $i\in\mathbb N$