Sum Challenge

[SANZEED]

Sum Challenge 001


Let the divisors of n be A={d1,d2,d3,….,dn} where
d1=1 and dn =n. Let Pd =∑i σ(di), d Є A, d/di.
Remember that σ(x)=sum of the divisors of x.

Prove that, I
∑d/n μ(d)Pd=1.

[Phlembac Adib Hasan]

Sum Challenge 001


Let the divisors of n be A={d1,d2,d3,….,dn} where
d1=1 and dn =n. Let Pd =∑i σ(di), d Є A, d/di.
Remember that σ(x)=sum of the divisors of x.

Prove that, I
∑d/n μ(d)Pd=1.

The problem statement is confusing. Please clearly define $P_d$.Is it $P_{d_i} $? Otherwise it seems that $ P_d $ has only one value for every $n$.If you can't use LaTeX, then please explain it.

[nafistiham]

i was requested by SANZEED to latex his problem.unfortunately, now i myself don't understand what he wants to ask

[SANZEED]

YES,$p_{d}$ means $p_{d_{i}}$.let $d$ be adivisor of n.Then take the value of the divisors of n for $d_{i}$ which are divisible by $d$.

Moderation Note: $L^AT_EX$ed correctly.

[Phlembac Adib Hasan]

বড় সংখ্যার জন্য কাজ করতেসে না। ক্যালকুলেশনেই ভুল করলাম কিনা

[SANZEED]

বড় সংখ্যার জন্য কাজ করতেসে না। ক্যালকুলেশনেই ভুল করলাম কিনা

ADB! you should try vinogradov's theorem

[SANZEED]

Have you found errors?But I checked my use of Vinogradov's......now Iam also confused....