Determine the number of solutions of the congruence x4≡61(mod 117)

Step 1 Given: x4≡61(mod 117) 117=32×13 As (ϕ(9))/(4,ϕ(9))=6/(4.6)=6/2=3(here(4.6)denotes the g.c.d of(4.6)) and (61)3≡(−2)3≡1(mod 9) we deduce the congruence x4≡61(mod 9) has (4,ϕ(9))=(4.6)=2 solutions Step 2 Similarity yϕ(13)4,ϕ(13)=124.12=124=3 and (61)3≡(−4)3≡1(mod 13) So, the congruence x4≡61(mod 13) has (4,ϕ(13))=(4.12)=4 solutions hence, the number of solutions of the congruence x4≡61(mod 117) is 2×4=8.

Question: "Determine the number of solutions of the congruence..."

High School
Geometry
High school geometry
Congruence

Tyra

Answered question

2020-12-25

Determine the number of solutions of the congruence

x^{4} \equiv 61 (m o d 117)

Answer & Explanation

Pohanginah

Skilled2020-12-26Added 96 answers

Step 1
Given: $x^{4} \equiv 61 (m o d 117)$
$117 = 3^{2} \times 13$
As $(ϕ (9)) / (4, ϕ (9)) = 6 / (4.6) = 6 / 2 = 3 (h e r e (4.6) d e n o t e s t h e g . c . d o f (4.6))$
and $(61)^{3} \equiv (- 2)^{3} \equiv 1 (m o d 9)$
we deduce the congruence
$x^{4} \equiv 61 (m o d 9) h a s (4, ϕ (9)) = (4.6) = 2 s o l u t i o n s$
Step 2
Similarity $y \frac{ϕ (13)}{4, ϕ (13)} = \frac{12}{4.12} = \frac{12}{4} = 3$
and $(61)^{3} \equiv (- 4)^{3} \equiv 1 (m o d 13)$
So, the congruence $x^{4} \equiv 61 (m o d 13) h a s (4, ϕ (13)) = (4.12) = 4 s o l u t i o n s$
hence, the number of solutions of the congruence $x^{4} \equiv 61 (m o d 117) i s 2 \times 4 = 8$ .

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