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I have been trying to evaluate $\sum_{i=1}^{125} i^4\pmod {125}$. My attempt has been somewhat like this so far:

We know that 125 has a primitive root. Let's call it $r$. Now we know that $$r,r^2,\ldots, r^{\phi(125)}$$ is actually congruent to the set of positive integers that are less than 125 and relatively prime to it, i.e. all the numbers not divisible by 5. Also note that $\phi(125)=100$. Thus I write the sum as follows: $$\sum_{i=1}^{125} i^4=(5^4+10^4+15^4+\ldots + 125 ^4)+(1+2^4+3^4+\ldots+124^4)$$ $$=5^4(1+2^4+3^4+\ldots+25^4)+(1+r^4+r^8+\ldots+r^{396})$$ $$\equiv \frac{r^{400}-1}{r^4-1}\pmod {125}$$ $$\equiv 0\pmod {125}$$

But when I calculate the expression using wolphram alpha, I get the answer is 100. Where am I going wrong? Please point out. Thanks in advance.

Jyrki Lahtonen
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Takamoto Yuji
  • 749

3 Answers3

1

There is not much number theory needed to evaluate this sum. Note that $$\sum_{i=0}^{125}i^4 = \sum_{i=0}^4\sum_{j=0}^4\sum _{k=0}^4(25i+5j+k)^4$$ Now we apply the binomial formula to $$(5(5i+j)+k)^4$$ and get $$5^4(5i+j)^4+{4\choose1}5^3(5i+j)^3k+{4\choose2}5^2(5i+j)^2k^2+{4\choose3}5(5i+j)k^3+k^4$$ The 1st term is $0\pmod{125}$ and if we sum over $k$ the 2nd, 3rd and 4th term vanishes, too, because $$\sum_{k=0}^4 k=2\cdot 5$$ $$\sum_{k=0}^4 k^2 =6\cdot 5$$ $$\sum_{k=0}^4 k^3 =4\cdot 5^2$$ so only $$\sum_{i=0}^4\sum_{j=0}^4\sum _{k=0}^4k^4$$ remains and is $$\sum_{i=0}^4 1 \sum_{j=0}^4 1 \sum_{i=0}^4 k^3\equiv100\pmod{125}$$

miracle173
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The problem in your attempt is that $r^4-1$ is divisible by five, and division by five is not well defined in the ring of residue classes modulo $125$. For example $125/5$ and $0/5$ are not congruent modulo $125$ even though $125$ and $0$ are. Therefore your sum formula is not valid. Do observe that the error will be a multiple of $25$ (because $\gcd(125,r^4-1)=5$ and $125/5=25$). This fits with the result you got from WA!

Instead, you should notice that:

  • If $i\equiv0\pmod5$, then $i^4\equiv0\pmod{125}$, and you can leave those terms out of the reckoning (you already did, actually).
  • In the cyclic group $G=\Bbb{Z}_{125}^*$ of order $100$, the fourth powers form a unique subgroup $H$ of index four (this happens in all cyclic groups of order divisible by four). We can easily identify that the subgroup consisting of residue classes of integers $\equiv1\pmod5$ is also such a subgroup, so $$H=\{\overline{a}\in\Bbb{Z}_{125}\mid a\equiv1\pmod5\}.$$
  • Raising to fourth power is a 4-to-1 mapping in $G$, implying (by basic properties of homomorphisms) that each element of $H$ is attained as $i^4$ exactly four times.
  • Therefore your task is to calculate the sum $$ \sum_{i=1}^{125}i^4\equiv4\sum_{x\in H}x=4\sum_{i=1, i\equiv1\pmod5}^{125}i. $$ This is the sum of an arithmetic progression, and I'm sure you can manage. You can calculate it as an integer, and then reduce modulo $125$.
Jyrki Lahtonen
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As $2^2\equiv-1\pmod5,2$ is a primitive root of $5$

Now from Order of numbers modulo $p^2$

ord$_{5^2}2=4$ or $4\cdot5 $

Now as $2^4\not\equiv1\pmod{25},$ord$_{5^2}2=4\cdot5=\phi(25)$

So, $2$ is a primitive root of $5^2$

Using If $g$ is a primitive root of $p^2$ where $p$ is an odd prime, why is $g$ a primitive root of $p^k$ for any $k \geq 1$?,

$2$ is a primitive root of $5^n, n\ge1$

Now $$\sum_{n=0}^{99}2^{4n}=\dfrac{16^{100}-1}{16-1}$$

Now $16^{100}\equiv?\pmod{125(16-1)}$

Now as $125\cdot15=5^4\cdot3, \displaystyle16^{100}=(1+15)^{100}\equiv1+\binom{100}115+\binom{100}215^2\pmod{5^4\cdot3}$

As $\displaystyle(15^2,125\cdot15)=75,\binom{100}2\equiv0\pmod{25}$

$\displaystyle\implies16^{100}-1\equiv1500\pmod{125\cdot15}$

$\displaystyle\implies\dfrac{16^{100}-1}{16-1}\equiv\dfrac{1500}{16-1}\pmod{125}$

lab bhattacharjee
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