Solve the following equation in positive numbers. $$(2a+1)(2a^2+2a+1)(2a^4+4a^3+6a^2+4a+1)=828567056280801$$
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Tags: algebra
23.11.2020 22:23
Expanding gives $$8 a^7 + 28 a^6 + 56 a^5 + 70 a^4 + 56 a^3 + 28 a^2 + 8 a + 1 = 828567056280801$$which is a binomial to the power of $8$ minice $a^8$. Adding $a^8$ to both sides gives $(a+1)^8 = 828567056280801 + a^8$. I quickly found that $\sqrt[8]{828567056280801^{(1/7+1/8)/2}} = 99.538...$. By checking, $100$ is an answer. I could not find any others REAL solutions.
23.11.2020 22:37
Jc426 wrote: I quickly found that $\sqrt[8]{828567056280801^{(1/7+1/8)/2}} = 99.538...$. By hand?
23.11.2020 22:47
no, lol. im not that bashy. by calculator
23.11.2020 22:52
You could tangent line approx I think but that is still annoying
23.11.2020 22:57
It is $8*\xi^7$ for $\xi\in [a,a+1]$, then $\xi^7\simeq 10^{14}\implies \xi$ is around 100.
23.11.2020 23:24
arqady wrote: Solve the following equation in positive numbers. $$(2a+1)(2a^2+2a+1)(2a^4+4a^3+6a^2+4a+1)=828567056280801$$ Let $x = a+1$ and $y=a$, so $x-y=1$. Since the terms are $x+y$, $x^2+y^2$, $x^4+y^4$ and $x-y=1$ we have \[ x^8-y^8 = (x-y)(x+y)(x^2+y^2)(x^4+y^4) = 828567056280801. \]On the other hand, the RHS is $\sum_{k=0}^7 100^k \binom{8}{k} = 101^8 - 100^8$, because the digits (taken in pairs) are binomial coefficients $\tbinom 8k$. Hence $a = 100$ is the unique positive solution.
23.11.2020 23:34
whoa just whoa
24.11.2020 18:52
v_Enhance wrote: arqady wrote: Solve the following equation in positive numbers. $$(2a+1)(2a^2+2a+1)(2a^4+4a^3+6a^2+4a+1)=828567056280801$$ Let $x = a+1$ and $y=a$, so $x-y=1$. Since the terms are $x+y$, $x^2+y^2$, $x^4+y^4$ and $x-y=1$ we have \[ x^8-y^8 = (x-y)(x+y)(x^2+y^2)(x^4+y^4) = 828567056280801. \]On the other hand, the RHS is $\sum_{k=0}^7 100^k \binom{8}{k} = 101^8 - 100^8$, because the digits (taken in pairs) are binomial coefficients $\tbinom 8k$. Hence $a = 100$ is the unique positive solution. That was also my solution in the contest. This problem is a troll, took me one hour. I thought this is number theory problem and that there is no solution, and i proved that $a$ must be of the form $3k+1$. Then the supervisor told me that $a$ is real number and doesn't have to be an integer (so there must be such $a$), so i realized that $a$ is some number that will be easy to do computations with