Exercise 5.8.5 (Repeated root of $x^3 + Ax+B$)

Show that if p > 3 , 4 A 3 + 27 B 2 ≡ 0 ( mod p ) , p ∤ A , then the root r of the congruence − 2 Ar ≡ 3 B ( mod p ) is a repeated root ( mod p ) of the polynomial x 3 + Ax + B (*).

Typo corrected (note of R.G.).

Answers

Proof. Let a = Ā ≠ 0 , b = B ¯ denote the classes of A , B in ℤ ∕ pℤ , where p > 3 .

If r ¯ ∈ ℤ ∕ pℤ is a repeated root of f ( x ) = x 3 + ax + b ∈ ( ℤ ∕ pℤ ) [ x ] , then f ( r ¯ ) = f ′ ( r ¯ ) = 0 . So

f ( r ¯ ) = r ¯ 3 + a r ¯ + b = 0 , (1) f ′ ( r ¯ ) = 3 r ¯ 2 + a = 0 . (2)

Then

0 = − 3 f ( r ¯ ) + r ¯ f ′ ( r ¯ ) = − 2 a r ¯ − 3 b . (3)

Thus r ¯ = − 3 b ( 2 a ) − 1 , and by (2), 3 [ − 3 ( b ( 2 a ) − 1 ] 2 + a = 0 , so

4 a 3 + 27 b 2 = 0 . (4)

Hence

4 A 3 + 27 B 2 ≡ 0 mod p , − 2 Ar ≡ 3 B ( mod p ) .

Conversely, suppose that 4 A 3 + 27 B 2 ≡ 0 ( mod p ) and − 2 Ar ≡ 3 B ( mod p ) , then

4 a 3 + 27 b 2 = 0 , − 2 a r ¯ = 3 b .

Then r ¯ = − ( 3 b ) ( 2 a ) − 1 , thus

f ′ ( r ¯ ) = 3 r ¯ 2 + a = 3 [ − ( 3 b ) ( 2 a ) − 1 ] 2 + a = ( 4 a 3 + 27 b 2 ) ( 2 a ) − 2 = 0 . f ( r ¯ ) = r ¯ 3 + a r ¯ + b = [ − ( 3 b ) ( 2 a ) − 1 ] 3 + a [ − ( 3 b ) ( 2 a ) − 1 ] + b = ( 2 a ) − 3 [ − ( 3 b ) 3 − a ( 3 b ) ( 2 a ) 2 + b ( 2 a ) 3 ] = − ( 2 a ) − 3 b [ 4 a 3 + 27 b 2 ] = 0 .

Since f ( r ¯ ) = f ′ ( r ¯ ) = 0 , r ¯ is a repeated root of x 3 + ax + b .

In conclusion, 4 A 3 + 27 B 2 ≡ 0 ( mod p ) , p ∤ A , p > 3 , then the root r of the congruence − 2 Ar ≡ 3 B ( mod p ) is a repeated root ( mod p ) of the polynomial x 3 + Ax + B .

(Moreover, by the first part, if x 3 + Ax + B , p ∤ A has a repeated root r ( mod p ) , where p > 3 , then 4 A 3 + 27 B 2 ≡ 0 ( mod p ) and − 2 Ar ≡ 3 B ( mod p ) .) □

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2025-07-01 08:41
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