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Original Question: Solve for all positive integers $n$ such that $25^n + 16^n \equiv 1 \pmod{121}$.

I began by substituting $k=2n$ to obtain $$5^k + 4^k \equiv 1 \pmod{11}$$

Considering this modulo $11$, it can be found that $k \equiv 4 \pmod{5}$. Then, I observed that $5^3 \equiv 4 \pmod{121}$, and so I split the congruence into three cases:

Case-1

$k \equiv 0 \pmod{3}$. Then, $k=3r$. $$(4^k)^3+4^k \equiv 1 \pmod{121}$$ It follows that $4^k \equiv 64 \pmod{121}$.

Case-2

$k \equiv 1 \pmod{3}$. Then, $k=3r+1$. $$4 \cdot (4^r)^3 +5 \cdot (4^r) \equiv 1 \pmod{121}$$ It follows that $4^k \equiv 37 \pmod{121}$.

Case-3

$$k \equiv 2 \pmod{3}.$$ Then,$$ k=3r+2.$$ $$16 \cdot (4^r)^3 +25 \cdot (4^r) \equiv 1 \pmod{121}.$$ It follows that $4^r \equiv 80 \pmod{121}$.

This leaves the congruences:

  • $4^k \equiv 64 \pmod{121}$ for $k \equiv 4 \pmod{15}$
  • $4^k \equiv 37 \pmod{121}$ for $k \equiv 9 \pmod{15}$
  • $4^k \equiv 80 \pmod{121}$ for $k \equiv 14 \pmod{15}$

From here, I am unsure how to proceed.

Indecisive
  • 631

2 Answers2

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Note that $121=11^2$, so the Euler indicator function $\varphi$ for $11^2$ is $$ \varphi(121)= \varphi(11^2)= \left(1-\frac 1{11}\right)\cdot 11^2=10\cdot 11=110\ . $$ The numbers $16=4^2$, $25=5^2$ are relatively prime to $121$, so their multiplicative orders modulo $121$ are divisors of $55$, since they are already squares. So $n_0$ is a solution if and only if all $n$ of the shape $n_0+55k$ are solutions.

The congruence $$ 25^n+16^n\equiv 1\qquad\text{ modulo }11 $$ has - as mentioned - solutions of the shape $n\equiv 2$ modulo $5$.

It is thus sufficient to check all $n$ from $0$ to $54$, and among them only those which are two modulo five. So we check the eleven numbers $n$ among $2, 7,12,17,22,27,32,37,42,47, 52$. For them we compute, with equality (considered in the ring $\Bbb Z/121$) written instead of equivalence (in $\Bbb Z$) some powers, and sums of them. Because of $25^{11}=3$, $16^{11}=3^3$, it is useful to observe that $3$ has multiplicative order five, $3^5=243=242+1=1$, to the powers of $3$ are $3^0=1$, $3^1=3$, $3^2=9$, $3^3=27$, $3^4=81=-40$. We compute:
$$ \begin{aligned} 5^3 &=125 =4=2^2\ ,\\ 2^7 &= 128 =7\ ,\\[3mm] 25^3 &= 5^6 = (5^3)^2=125^2=4^2=16\ ,\\ 16^{37} &= (25^3)^{37}=25^{111}=25^{110+1}=25^{\varphi(121)+1}=25\ ,\\ 25^{11} &= 5^{22}=(5^3)^7\cdot 5=125^7\cdot 5=4^7\cdot 5=2^{14}\cdot 5 \\ &=(2^7)^2\cdot 5 = 128^2\cdot 5=7^2\cdot 5=49\cdot 5=245 \\ &=3\ , \\ 16^{11} &=(2^4)^{11}=2^{44}=(2^7)^6\cdot 4=128^6\cdot 4=7^6\cdot 4=343^2\cdot 4=(-20)^2\cdot 4 \\ &=1600=390=27=3^3\ , \\ 25^2 &= 625 = 20\ ,\\ 16^2 &= 256 = 14\ , \\[4mm] 25^2 + 16^2 &= 20 + 14 = 34\ne 1\ ,\\ 25^7 + 16^7 &= (5^3)^4\cdot 25 +(2^7)^4 = 4^4\cdot 25+7^4 =7\cdot 2\cdot 25+7\cdot 343 = 7(50-20)=210=89\ne 1\ ,\\ 25^{12} + 16^{12} &= 25^{11}\cdot 25+16^{11}\cdot 16=3\cdot 25 + 27\cdot 16 =75+432=507=484+23=23\ne 1\ ,\\ 25^{17} + 16^{17} &= 25^{11}\cdot (5^3)^4 + 16^{11}\cdot (2^7)^3\cdot 2^3 =3\cdot4^4 + 27\cdot 7^3\cdot 8=6\cdot 2^7+27\cdot(-20)\cdot 8 \\ &=42-(27\cdot 10)\cdot 16=42-28\cdot 16=-406=484-406=78\ne 1\ , \\ 25^{22} + 16^{22} &= (25^{11})^2 + (16^{11})^2=3^2+(3^3)^2 =3^2+3^5\cdot 3=3^2+3=12\ne 1\ , \\ 25^{27} + 16^{27} &= (25^{11})^2\cdot (5^3)^3\cdot 5 + (16^{11})^2\cdot (2^7)^2\cdot 2^6 =3^2\cdot 4^3\cdot 5+(3^3)^2\cdot 7^2\cdot 2^6 \\ &=3\cdot 8\cdot 120 +3^5\cdot 3\cdot 49\cdot 2^6 =-24+1\cdot 147\cdot 64=-24+26\cdot2^6 \\ &=-24+13\cdot 7=67 \ne 1\ , \\ \color{green}{ 25^{32} + 16^{32}} &\color{green}{= (25^{11})^2\cdot (5^3)^6\cdot 5^2 + (16^{11})^2\cdot (2^7)^5\cdot 2^5 =3^2\cdot4^6\cdot 25+3^6\cdot 7^5\cdot 2^5} \\ &\color{green}{=225\cdot 2^7\cdot 2^5+3^5\cdot 3\cdot 7^2\cdot343\cdot 2^5 ((-17)\cdot 7 +1\cdot 147\cdot(-20))2^5} \\ &\color{green}{=(2-36)\cdot 32=-32\cdot 34=-(33-1)(33+1)=-33^2+1=1}\ , \\ 25^{37} + 16^{37} &= 25^{37} +1\ne 1\ , \\ 25^{42} + 16^{42} &= (25^{11})^3\cdot 125^6 +16^{37}\cdot 16^5 =3^3\cdot 4^6+25\cdot (2^7)^2\cdot 2^6 =27\cdot 2^7\cdot 2^5+25\cdot 7^2\cdot 8\cdot 8 \\ &=27\cdot 7\cdot 32 + 200\cdot 392 =27\cdot(-18) +(-42)\cdot 29=-2\cdot 3^5-1218 \\ &=-2-8=-10=111\ne 1\ , \\ 25^{47} + 16^{47} &= (25^{11})^4\cdot 125^2 +16^{37}\cdot 16^{10} =3^4\cdot 4^2+25\cdot (2^7)^5\cdot 2^5 \\ &=1296+ 25\cdot4\cdot 7^3\cdot 49\cdot 8 =86+100\cdot(-20)\cdot 392 =86+(-21)(-20)\cdot 29 \\ &=12266=166=45\ne 1\ , \\ 25^{52} + 16^{52} &= (25^{11})^4\cdot 5^{16} +16^{37}\cdot 16^{11}\cdot 16^4&= 3^4\cdot(5^3)^5\cdot 5+25\cdot 3^3\cdot (2^7)^2\cdot 4 =3^3(3\cdot 4^5\cdot 5+25\cdot 7^2\cdot 4) \\ &=3^3(120\cdot 2^7+4900)=3^3(-7+60)=3^3\cdot 53=9\cdot 38=100\ne 1\ . \end{aligned} $$ The solution is thus: $$ \bbox[yellow]{ n\text{ is congruent to $32$ modulo $55$ .}} $$

dan_fulea
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As $5^3\equiv4\pmod{121},$

$$p(n)=5^{2n}+4^{2n}\equiv5^{2n}+(5^3)^{2n}\pmod{121}$$

Let's verify $5^{2n}+5^{6n}\pmod{11}$

As $5^2\equiv3\pmod{11},5^3\equiv15\equiv4,5^5\equiv3\cdot4\equiv1,$

$p(n)=5^{2n}+5^{6n}\equiv5^n(1+5^n)\pmod{11}$ and it will have a period of $5$

$p(1)\equiv5(1+6)\not\equiv1\pmod{11}$

$p(2)\equiv5^2(1+5^2)\equiv3(1+3)\equiv1\implies n=5m+2$ where $m$ is a non-negative integer

$5^3\equiv4\pmod{121},5^5\equiv5^2\cdot4\not\equiv1\pmod{11^2}$

Using Order of numbers modulo $p^2$, modulo ord$_{121}5=11\cdot5\ \ \ \ (1)$

$p(5m+2)=5^{2(5m+2)}+4^{2(5m+2)}\equiv5^4\cdot(5^{10})^m+4^4\cdot(4^{10})^m$

$5^4\equiv5\cdot4\pmod{121},5^5\equiv4\cdot5^2\equiv-21,5^{10}\equiv(-21)^2\equiv1+77$

As using binomial theorem $(1+77)^m=1+\binom m177^1+$terms divisible by $77^2,$

$\implies5^4\cdot(5^{10})^m\equiv20(1+77)^m\equiv20(1+77m)\pmod{121}\equiv20+88m$

Similarly, $4^4\cdot(4^{10})^m\equiv14(1-11)^m\equiv14(1-11m)\pmod{121}\equiv14-33m$

$\implies p(5m+2)\equiv34+55m$

So, we need $$34+55m\equiv1\pmod{121}\iff55m\equiv-33\pmod{121}\iff5m\equiv-3\pmod{11}$$

As $5\cdot2\equiv-1\pmod{11}\implies5^{-1}\equiv-2$

$\implies m\equiv(-3)(-2)\pmod{11}\implies m=11r+6$

$\implies n=5(11r+6)+2\equiv32\pmod{55}$ by $(1)$

lab bhattacharjee
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