Concepts in Abstract Mathematics

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Concepts in Abstract Mathematics

Homework questions – Week 5

Jean-Baptiste Campesato

February 22^nd, 2021 to February 26^th, 2021

Exercise 1.

1. Prove that∀𝑛 ∈ ℕ, 5|2^2𝑛+1+ 3^2𝑛+1

2. Prove that∀𝑛 ∈ ℕ, 17|2^7𝑛+1+ 3^2𝑛+1+ 5^10𝑛+1+ 7^6𝑛+1 Exercise 2.

Find all the𝑥 ∈ ℤsuch that𝑥²+ 3 ≡ 0 (mod7).

Exercise 3.

1. Determine the remainder of the Euclidean division of2^𝑛by5for𝑛 ∈ ℕ.

2. Determine the remainder of1357²⁰²¹by5.

Exercise 4.

1. Find a criterion for divisibility by5.

Use it on958547and on123456789336.

Use it on123456789and715.

Exercise 5.

1. Find the integer solutions of𝑥²− 5𝑦²= 3.

2. Find the integer solutions of15𝑥²− 7𝑦²= 9.

3. Find the integer solutions of𝑥²+ 𝑦² = 4003 (Hint: work modulo 4).

Exercise 6.

Prove that13|3¹²⁶+ 5¹²⁶. Exercise 7.

• For which𝑛 ∈ ℕ, is it true that8|3^𝑛+ 4𝑛 + 1?

• For which𝑛 ∈ ℕ, is it true that21|2²^𝑛+ 2^𝑛+ 1?

Exercise 8.

1. Prove that∀𝑎, 𝑏 ∈ ℤ, (3|𝑎and3|𝑏) ⇔ 3|(𝑎²+ 𝑏²).

2. Prove that∀𝑎, 𝑏 ∈ ℤ, (7|𝑎and7|𝑏) ⇔ 7|(𝑎²+ 𝑏²).

3. Prove that∀𝑎, 𝑏 ∈ ℤ, 21|(𝑎²+ 𝑏²) ⟹ 441|(𝑎²+ 𝑏²).

Exercise 9.

Compute gcd(2⁴⁴⁵+ 7, 15). Exercise 10.

Find all the prime numbers𝑝such that2^𝑝+ 𝑝²is also prime.

Exercise 11.

What is the last digit in the decimal expansion of7³⁸⁴?

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1. Convert the following number from the Babylonian cuneiform numeral system to base10:

𒄴𒐌 𒄭𒈫 𒐈 𒌋 𒁹

2. Convert the following number from decimal to the Babylonian cuneiform numeral system:42137.

3. Convert the following number from hexadecimal (with digits0, 1, 2, … , 9, 𝐴, 𝐵, … , 𝐹) to base10:𝐹 420𝐶¹⁶. 4. Convert the following number from decimal to hexadecimal: 11211.

5. Compute in hexadecimal (without converting to decimal):9𝐴𝐵7¹⁶+ 3𝐶0𝐷¹⁶. 6. Perform the above computation using decimals. Is it easier?

Exercise 13.

The scene takes place on an island inhabited by chameleons which are either blue, green, or red.

When two chameleons of different colors meet, they both change to the third color (for instance, if a green chameleon and a red chameleon meet, then they both become blue).

Cherge, one of the chameleons, is a retired mathematician who likes funny mathematical riddles and tongue twisters. While he stands at the highest place on the island, he is able to see all the chameleons: 17 of them are blue, 15 are green and 13 are red (including himself).

Then he wonders about a new riddle”Could the island become monochromatic?”. What do you think?

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Sample solutions to Exercise 1.

1. Note that2²= 4 ≡ −1 (mod5)and that3² = 9 ≡ −1 (mod5). Therefore, for𝑛 ∈ ℕ, we have

2^2𝑛+1+ 3^2𝑛+1= (2²)^𝑛× 2 + (3²)^𝑛× 3 ≡ (−1)^𝑛× 2 + (−1)^𝑛× 3 (mod5) ≡ (−1)^𝑛× 5 (mod5) ≡ 0 (mod5).

2. Let𝑛 ∈ ℕ, then

2^7𝑛+1+ 3^2𝑛+1+ 5^10𝑛+1+ 7^6𝑛+1≡ 9^𝑛× 2 + 9^𝑛× 3 + 9^𝑛× 5 + 9^𝑛× 7 (mod17) ≡ 9^𝑛× 17 (mod17) ≡ 0 (mod17) Sample solutions to Exercise 2.

We first compute𝑥²+ 3 (mod7)in terms of𝑥 (mod7):

𝑥 (mod7) 0 1 2 3 4 5 6 𝑥²(mod7) 0 1 4 2 2 4 1 𝑥²+ 3 (mod7) 3 4 0 5 5 0 4

Let𝑥 ∈ ℤ. Then𝑥²+ 3 ≡ 0 (mod7)if and only if𝑥 ≡ 2 (mod7)or𝑥 ≡ 5 (mod7) if and only if𝑥 ∈ {2 + 7𝑘 ∶ 𝑘 ∈ ℤ} ∪ {5 + 7𝑘 ∶ 𝑘 ∈ ℤ}.

1. We first look for the least𝑘 ∈ ℕ ⧵ {0}such that2^𝑘≡ 1 (mod5):

• 2¹≡ 2 (mod5)

• 2²≡ 4 (mod5)

• 2³≡ 3 (mod5)

• 2⁴≡ 1 (mod5) Hence it is4.

We perform the Euclidean division of𝑛 ∈ ℕby4:𝑛 = 4𝑞 + 𝑟where0 ≤ 𝑟 < 4.

Then2^𝑛= 2^4𝑞+𝑟= (2⁴)^𝑞2^𝑟≡ 1^𝑞2^𝑟(mod5) ≡ 2^𝑟(mod5).

Thus

• If𝑛 ≡ 0 (mod4)then2^𝑛≡ 2⁰(mod5) ≡ 1 (mod5), so the remainder is1.

• If𝑛 ≡ 1 (mod4)then2^𝑛≡ 2¹(mod5) ≡ 2 (mod5), so the remainder is2.

• If𝑛 ≡ 2 (mod4)then2^𝑛≡ 2²(mod5) ≡ 4 (mod5), so the remainder is4.

• If𝑛 ≡ 3 (mod4)then2^𝑛≡ 2³(mod5) ≡ 3 (mod5), so the remainder is3.

2. Note that1357 = 1355 + 2 ≡ 2 (mod5). Therefore1357²⁰²¹≡ 2²⁰²¹(mod5).

Since2021 = 505 × 4 + 1 ≡ 1 (mod4)we get that the remainder of1357²⁰²¹by5is2.

1. Note that10 ≡ 0 (mod5), hence

5|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ ⇔ 𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ ≡ 0 (mod5)

⇔

𝑟

∑𝑘=0

𝑎_𝑘10^𝑘≡ 0 (mod5)

⇔ 𝑎₀≡ 0 (mod5)

Therefore5|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰if and only if𝑎₀ = 0or𝑎₀ = 5.

2. Note that10³= 175 × 8 ≡ 0 (mod8), hence 8|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ ⇔ 𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ ≡ 0 (mod8)

⇔

𝑟

∑𝑘=0

𝑎_𝑘10^𝑘≡ 0 (mod8)

⇔ 10²𝑎₂+ 10𝑎₁+ 𝑎₀≡ 0 (mod8)

⇔ 4𝑎₂+ 2𝑎₁+ 𝑎₀ ≡ 0 (mod8)

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Therefore8|𝑎_𝑟𝑎_𝑟−1… 𝑎₀ if and only if8|(4𝑎₂+ 2𝑎₁+ 𝑎₀).

Note that8|958547if and only if8|4 × 5 + 2 × 4 + 7 = 35 = 8 × 4 + 3. Therefore8 ∤ 958547.

Note that8|123456789336if and only if8|4 × 3 + 2 × 3 + 6 = 24 = 8 × 3. Therefore8|123456789336.

3. Note that10 ≡ −1 (mod11), hence

11|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰⇔ 𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰ ≡ 0 (mod11)

⇔

𝑟

∑𝑘=0

𝑎_𝑘10^𝑘≡ 0 (mod11)

⇔

𝑟

∑𝑘=0

(−1)^𝑘𝑎_𝑘≡ 0 (mod11)

Therefore11|𝑎_𝑟𝑎_𝑟−1… 𝑎₀¹⁰if and only if11|(−1)^𝑟𝑎_𝑟+ (−1)^𝑟−1𝑎_𝑟−1+ ⋯ + 𝑎₂− 𝑎₁+ 𝑎₀.

Note that11|123456789if and only if11|9 − 8 + 7 − 6 + 5 − 4 + 3 − 2 + 1 = 5. Therefore118 ∤ 123456789.

Note that11|715if and only if11|5 − 1 + 7 = 11. Therefore11|715.

1. If(𝑥, 𝑦) ∈ ℤ²is a solution then𝑥² ≡ 3 (mod5). But

• if𝑥 ≡ 0 (mod5)then𝑥²≡ 0 (mod5),

• if𝑥 ≡ ±1 (mod5)then𝑥²≡ 1 (mod5),

• if𝑥 ≡ ±2 (mod5)then𝑥²≡ 4 (mod5).

Thus the equation has no integer solution.

2. Assume that(𝑥, 𝑦) ∈ ℤ²is a solution, then taking congruences modulo 3, the equation becomes 0𝑥²− (−1)𝑦² ≡ 0 (mod3)

i.e.𝑦²≡ 0 (mod3).

𝑦 (mod3) 0 1 2 𝑦²(mod3) 0 1 1

Therefore𝑦 ≡ 0 (mod3), i.e. 𝑦 = 3𝑘for some𝑘 ∈ ℤ, and the equation becomes15𝑥²− 63𝑘² = 9.

Dividing by3, we get5𝑥²− 21𝑘² = 3. Taking congruences modulo 3, we obtain−𝑥²≡ 0 (mod3).

As above, the only possibility is that𝑥 ≡ 0 (mod3), i.e. 𝑥 = 3𝑙for some𝑙 ∈ ℤ.

Then the equation becomes45𝑙²− 21𝑘² = 3, and dividing by3, we get15𝑙²− 7𝑘²= 1.

Modulo 3, we finally get−𝑘²≡ 1 (mod3), i.e.𝑘² ≡ −1 (mod3) ≡ 2 (mod3).

Which is impossible (a square modulo 3 is either congruent to 0 or 1, according to the above array).

Thus the equation has no integer solution.

3. Below are the possible values for𝑥²(mod4)depending on𝑥 (mod4).

𝑥 (mod4) 0 1 2 3 𝑥²(mod4) 0 1 0 1

Therefore either𝑥² ≡ 0 (mod4)or𝑥²≡ 1 (mod4)and similarly either𝑦² ≡ 0 (mod4)or𝑦²≡ 1 (mod4).

Thus either𝑥²+ 𝑦²≡ 0 (mod4), or𝑥²+ 𝑦²≡ 1 (mod4), or𝑥²+ 𝑦²≡ 2 (mod4).

Since4003 = 4 × 1000 + 3 ≡ 3 (mod4), there is no integer solutions.

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Note that3³≡ 1 (mod13). Since126 = 3 × 42, we get3¹²⁶ ≡ (3³)⁴²(mod13) ≡ 1⁴²(mod13) ≡ 1 (mod13).

Note that5⁴ ≡ 1 (mod13). Since126 = 4 × 31 + 2, we get5¹²⁶ ≡ (5⁴)³¹× 5²(mod13) ≡ 1³¹× 25 (mod13) ≡

−1 (mod13).

Therefore3¹²⁶+ 5¹²⁶ ≡ 0 (mod13).

1. Let𝑛 ∈ ℕ.

• If𝑛is even, i.e. 𝑛 = 2𝑘, then3^𝑛+ 4𝑛 + 1 = 9^𝑘+ 8𝑘 + 1 ≡ 1^𝑘+ 0 + 1 (mod8) ≡ 2 (mod8).

• If𝑛is odd, i.e. 𝑛 = 2𝑘+1, then3^𝑛+4𝑛+1 = 9^𝑘×3+8𝑘+4+1 ≡ 1^𝑘×3+0+4+1 (mod8) ≡ 0 (mod8).

Therefore8|3^𝑛+ 4𝑛 + 1if and only if𝑛is odd.

2. Let𝑛 ∈ ℕ. Note that2⁶ = 64 = 21 × 3 + 1 ≡ 1 (mod21).

Therefore, if𝑛 = 6𝑞 + 𝑟with0 ≤ 𝑟 < 6, we have that2^𝑛= (2⁶)^𝑞× 2^𝑟≡ 2^𝑟(mod21).

Thus2^𝑛(mod21)depends only on𝑛 (mod6). Let’s study the cases separately.

𝑛 (mod6) 0 1 2 3 4 5

2^𝑛(mod21) 1 2 4 8 16 11 2²^𝑛(mod21) 2 4 −5 4 16 −10 2²^𝑛+ 2^𝑛+ 1 (mod21) 4 7 0 13 12 2 Therefore21|2²^𝑛+ 2^𝑛+ 1if and only if𝑛 ≡ 2 (mod6).

1. For𝑎, 𝑏 ∈ ℤ, we compute𝑎²+ 𝑏²(mod3)depending on𝑎 (mod3)and𝑏 (mod3):

𝑎 (mod3)

𝑏 (mod3)

0 1 2

0 0 1 1

1 1 2 2

2 1 2 2

We see that𝑎²+ 𝑏²≡ 0 (mod3)if and only if𝑎 ≡ 0 (mod3)and𝑏 ≡ 0 (mod3).

2. Same as above.

3. Let𝑎, 𝑏 ∈ ℤ. Assume that21|𝑎²+ 𝑏². Then3|𝑎²+ 𝑏², thus3|𝑎and3|𝑏from the first question.

Similarly7|𝑎and7|𝑏from the second question.

Therefore the least common divisor of3and7divides𝑎and𝑏, i.e.21|𝑎and21|𝑏.

Hence𝑎 = 21𝑘and𝑏 = 21𝑙, so𝑎²+ 𝑏²= 441(𝑘²+ 𝑙²).

Note that2⁴≡ 1 (mod1)5. Since445 = 4 × 111 + 1we get

2⁴⁴⁵+ 7 = (2⁴)¹¹¹× 2 + 7 ≡ 1¹¹¹× 2 + 7 (mod15) ≡ 9 (mod15) Therefore, there exists𝑘 ∈ ℤsuch that2⁴⁴⁵+ 7 = 15𝑘 + 9.

Finally gcd(2⁴⁴⁵+ 7, 15) =gcd(15𝑘 + 9, 15) =gcd(9, 15) = 3.

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Note that2doesn’t work and that3works.

Assume that𝑝is a prime number greater than3, then

2^𝑝+ 𝑝²≡ (−1)^𝑝+ (±1)²(mod3) ≡ −1 + 1 (mod3) ≡ 0 (mod3) so3|2^𝑝+ 𝑝²and thus2^𝑝+ 𝑝²is not prime.

The only prime number𝑝such that2^𝑝+ 𝑝²is also prime is𝑝 = 3.

Note that7²≡ −1 (mod10)so7⁴ ≡ 1 (mod10).

Therefore if𝑛 = 4𝑞 + 𝑟with0 ≤ 𝑟 < 4, we get that7^𝑛 = (7⁴)^𝑞× 7^𝑟 ≡ 1^𝑞× 7^𝑟(mod10) ≡ 7^𝑟(mod10).

Hence it is enough to compute3⁸⁴ (mod4). Note that3²= 9 ≡ 1 (mod4), therefore 3⁸⁴ = 3⁸³^×8= (3²)⁸³^×4≡ 1⁸³^×4(mod4) ≡ 1 (mod4)

and3⁸⁴ = 4𝑞 + 1for some𝑞 ∈ ℕ Therefore 7³⁸⁴ = 7^4𝑞+1 ≡ 7 (mod 10). So the last digit in the decimal expansion of7³⁸⁴ is7.

1. 𒄴𒐌 𒄭𒈫 𒐈 𒌋 𒁹 = 57 × 60³+ 42 × 60²+ 3 × 60 + 11 × 1 = 12463391 2. We perform successive Euclidean division by 60:

42137 = 702 × 60 + 17

= (11 × 60 + 42) × 60 + 17

= 11 × 60²+ 42 × 60 + 17

= 𒌋 𒁹 𒄭𒈫 𒌋 𒐌

3. 𝐹 42𝐶¹⁶= 15 × 16⁴+ 4 × 16³+ 2 × 16²+ 0 × 16 + 12 = 999948 4. We perform successive Euclidean division by 16:

11211 = 700 × 16 + 11

= (43 × 16 + 12) × 16 + 11

= ((2 × 16 + 11) × 16 + 12) × 16 + 11

= 2 × 16³+ 11 × 16²+ 12 × 16 + 11

= 2𝐵𝐶𝐵¹⁶

5.

1 1

9 AB 7 + 3 C 0 D

=D 6 C 4

6. 9𝐴𝐵7¹⁶= 9 × 16³+ 10 × 16²+ 11 × 16 + 7 = 39607 3𝐶0𝐷¹⁶= 3 × 16³+ 12 × 16²+ 0 × 16 + 13 = 15373 39607 + 15373 = 54980

58820 = 3436×16+4 = (214×16+12)×16+4 = ((13×16+6)×16+12)×16+4 = 13×16³+6×16²+12×16+4 Therefore9𝐴𝐵7¹⁶+ 3𝐶0𝐷¹⁶ = 𝐷6𝐶4¹⁶I think it is easier to directly compute in base 16!

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Let’s denote the number of blue, green, and red chameleons respectively by𝑏,𝑔and𝑟.

• If a blue and a green chameleons meet, the new repartition bescomes𝑏^′= 𝑏−1,𝑔^′= 𝑔 −1and𝑟^′= 𝑟+2.

Therefore𝑏^′− 𝑔^′= 𝑏 − 𝑔,𝑏^′− 𝑟^′ = 𝑏 − 𝑟 − 3and𝑔^′− 𝑟^′= 𝑔 − 𝑟 − 3.

• Similarly, if a blue and a red chameleons meet, we have𝑏^′= 𝑏 − 1,𝑔^′= 𝑔 + 2and𝑟^′ = 𝑟 − 1.

Therefore𝑏^′− 𝑔^′= 𝑏 − 𝑔 − 3,𝑏^′− 𝑟^′= 𝑏 − 𝑟and𝑔^′− 𝑟^′= 𝑔 − 𝑟 + 3.

• Finally, if a green and a red chameleons meet, we have𝑏^′= 𝑏 + 2,𝑔^′ = 𝑔 − 1and𝑟^′ = 𝑟 − 1.

Therefore𝑏^′− 𝑔^′= 𝑏 − 𝑔 + 3,𝑏^′− 𝑟^′= 𝑏 − 𝑟 + 3and𝑔^′− 𝑟^′ = 𝑔 − 𝑟.

Note that in all the cases we have

𝑏^′− 𝑔^′≡ 𝑏 − 𝑔 (mod3) 𝑏^′− 𝑟^′≡ 𝑏 − 𝑟 (mod3) 𝑔^′− 𝑟^′≡ 𝑔 − 𝑟 (mod3)

Therefore these three quantities modulo 3 don’t change when the chameleons meet, they always stay con- stant, mathematically we say that they are invariant.

At the beginning, we have

𝑏 − 𝑔 ≡ 2 (mod3) 𝑏 − 𝑟 ≡ 1 (mod3) 𝑔 − 𝑟 ≡ 2 (mod3)

Assume by contradiction that all the chameleons become blue after several meetings (i.e.𝑏 = 45,𝑔 = 0and 𝑟 = 0), then

𝑏 − 𝑔 ≡ 0 (mod3) 𝑏 − 𝑟 ≡ 0 (mod3) 𝑔 − 𝑟 ≡ 0 (mod3)

Since these quantities don’t change when chameleons meet, we obtain a contradiction. Therefore, it is not possible to obtain an island with only blue chameleons from the initial situation.

We conclude similarly for the other colors. Thus, it is not possible to obtain a monochromatic island!

”Suis-je bien chez ce cher Serge ?”