P=7 vì 24+1=25 nên sẽ ko có số \(P^2\)thỏa mãn nên:

24x2+1=49 mà  \(7^2\)=49 và 7 cũng là SNT nên P=7

tick nka

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bài này khó quá

P=3+2^2(2+1)+2^4(2+1)+2^6(2+1)

=3(1+2^2+2^4+2^6)

=>đpcm

p;q là các số nguyên tố lớn hơn5

=>p,q là các số lẻ và p,q đều không chia hết cho 3

p là số lẻ nên p=2a+1

\(p^4-1=\left(p^2-1\right)\left(p^2+1\right)\)

=(p-1)(p+1)\(\left(p^2+1\right)\)

\(=\left(2a+1-1\right)\left(2a+1+1\right)\left\lbrack\left(2a+1\right)^2+1\right\rbrack\)

\(=2a\left(2a+2\right)\left(4a^2+4a+1+1\right)=2a\cdot2\cdot\left(a+1\right)\cdot2\cdot\left(2a^2+2a+1\right)\)

\(=8a\left(a+1\right)\left(2a^2+2a+1\right)\) ⋮8

p không chia hết cho 3 nên p=3k+1 hoặc p=3k+2

TH1: p=3k+1

\(p^4-1=\left(p-1\right)\left(p+1\right)\left(p^2+1\right)\)

\(=\left(3k+1-1\right)\left(3k+1+1\right)\left\lbrack\left(3k+1\right)^2+1\right\rbrack\)

\(=3k\left(3k+2\right)\left(9k^2+6k+1+1\right)=3k\left(3k+2\right)\left(9k^2+6k+2\right)\) ⋮3(2)

TH2: p=3k+2

\(p^4-1=\left(p-1\right)\left(p+1\right)\left(p^2+1\right)\)

\(=\left(3k+2-1\right)\left(3k+2+1\right)\left\lbrack\left(3k+2\right)^2+1\right\rbrack\)

\(=\left(3k+1\right)\left(3k+3\right)\left(9k^2+12k+4+1\right)=3\left(k+1\right)\left(3k+1\right)\left(9k^2+12k+5\right)\) ⋮3(1)

Từ (1),(2) suy ra \(p^4-1\) ⋮3

p là số nguyên tố lớn hơn 5

=>p không chia hết cho 5

=>p∈{5k+1;5k+2;5k+3;5k+4}

TH1: p=5k+1

\(p^4-1=\left(p-1\right)\left(p+1\right)\left(p^2+1\right)\)

\(=\left(5k+1-1\right)\left(5k+1+1\right)\left\lbrack\left(5k+1\right)^2+1\right\rbrack\)

\(=5k\left(5k+2\right)\left(25k^2+10k+1+1\right)\)

\(=5k\left(5k+2\right)\left(25k^2+10k+2\right)\) ⋮5(3)

TH2: p=5k+2

\(p^4-1=\left(p-1\right)\left(p+1\right)\left(p^2+1\right)\)

\(=\left(5k+2-1\right)\left(5k+2+1\right)\left\lbrack\left(5k+2\right)^2+1\right\rbrack\)

\(=\left(5k+1\right)\left(5k+3\right)\left(25k^2+20k+4+1\right)\)

\(=\left(5k+1\right)\left(5k+3\right)\left(25k^2+20k+5\right)\)

\(=5\left(5k^2+4k+1\right)\left(5k+1\right)\left(5k+3\right)\) ⋮5(4)

TH3: p=5k+3

\(p^4-1=\left(p-1\right)\left(p+1\right)\left(p^2+1\right)\)

\(=\left(5k+3-1\right)\left(5k+3+1\right)\left\lbrack\left(5k+3\right)^2+1\right\rbrack\)

\(=\left(5k+2\right)\left(5k+4\right)\left(25k^2+30k+9+1\right)\)

\(=\left(5k+2\right)\left(5k+4\right)\left(25k^2+30k+10\right)\)

\(=5\left(5k^2+6k+2\right)\left(5k+2\right)\left(5k+4\right)\) ⋮5(5)

TH4: p=5k+4

\(p^4-1=\left(p-1\right)\left(p+1\right)\left(p^2+1\right)\)

\(=\left(5k+4-1\right)\left(5k+4+1\right)\left\lbrack\left(5k+4\right)^2+1\right\rbrack\)

\(=\left(5k+3\right)\left(5k+5\right)\left(25k^2+40k+16+1\right)\)

\(=5\left(k+1\right)\left(5k+3\right)\left(25k^2+40k+17\right)\) ⋮5(6)

Từ (3),(4),(5),(6) suy ra \(p^4-1\) ⋮5

mà \(p^4-1\) ⋮3 và \(p^4-1\) ⋮8; \(p^4-1\) ⋮2

và ƯCLN(3;5;8;2)=1

nên \(p^4-1\) ⋮3*5*8*2

=>\(p^4-1\) ⋮240(7)

q là số lẻ nên q=2b+1

\(q^4-1=\left(q^2-1\right)\left(q^2+1\right)\)

=(q-1)(q+1)\(\left(q^2+1\right)\)

\(=\left(2b+1-1\right)\left(2b+1+1\right)\left\lbrack\left(2b+1\right)^2+1\right\rbrack\)

\(=2b\left(2b+2\right)\left(4b^2+4b+1+1\right)=2b\cdot2\cdot\left(b+1\right)\cdot2\cdot\left(2b^2+2b+1\right)\)

\(=8b\left(b+1\right)\left(2b^2+2b+1\right)\) ⋮8

q không chia hết cho 3 nên p=3k+1 hoặc p=3k+2

TH1: q=3k+1

\(q^4-1=\left(q-1\right)\left(q+1\right)\left(q^2+1\right)\)

\(=\left(3k+1-1\right)\left(3k+1+1\right)\left\lbrack\left(3k+1\right)^2+1\right\rbrack\)

\(=3k\left(3k+2\right)\left(9k^2+6k+1+1\right)=3k\left(3k+2\right)\left(9k^2+6k+2\right)\) ⋮3(8)

TH2: q=3k+2

\(q^4-1=\left(q-1\right)\left(q+1\right)\left(q^2+1\right)\)

\(=\left(3k+2-1\right)\left(3k+2+1\right)\left\lbrack\left(3k+2\right)^2+1\right\rbrack\)

\(=\left(3k+1\right)\left(3k+3\right)\left(9k^2+12k+4+1\right)=3\left(k+1\right)\left(3k+1\right)\left(9k^2+12k+5\right)\) ⋮3(9)

Từ (8),(9) suy ra \(q^4-1\) ⋮3

q là số nguyên tố lớn hơn 5

=>q không chia hết cho 5

=>q∈{5k+1;5k+2;5k+3;5k+4}

TH1: q=5k+1

\(q^4-1=\left(q-1\right)\left(q+1\right)\left(q^2+1\right)\)

\(=\left(5k+1-1\right)\left(5k+1+1\right)\left\lbrack\left(5k+1\right)^2+1\right\rbrack\)

\(=5k\left(5k+2\right)\left(25k^2+10k+1+1\right)\)

\(=5k\left(5k+2\right)\left(25k^2+10k+2\right)\) ⋮5(10)

TH2: q=5k+2

\(q^4-1=\left(q-1\right)\left(q+1\right)\left(q^2+1\right)\)

\(=\left(5k+2-1\right)\left(5k+2+1\right)\left\lbrack\left(5k+2\right)^2+1\right\rbrack\)

\(=\left(5k+1\right)\left(5k+3\right)\left(25k^2+20k+4+1\right)\)

\(=\left(5k+1\right)\left(5k+3\right)\left(25k^2+20k+5\right)\)

\(=5\left(5k^2+4k+1\right)\left(5k+1\right)\left(5k+3\right)\) ⋮5(11)

TH3: q=5k+3

\(q^4-1=\left(q-1\right)\left(q+1\right)\left(q^2+1\right)\)

\(=\left(5k+3-1\right)\left(5k+3+1\right)\left\lbrack\left(5k+3\right)^2+1\right\rbrack\)

\(=\left(5k+2\right)\left(5k+4\right)\left(25k^2+30k+9+1\right)\)

\(=\left(5k+2\right)\left(5k+4\right)\left(25k^2+30k+10\right)\)

\(=5\left(5k^2+6k+2\right)\left(5k+2\right)\left(5k+4\right)\) ⋮5(12)

TH4: q=5k+4

\(q^4-1=\left(q-1\right)\left(q+1\right)\left(q^2+1\right)\)

\(=\left(5k+4-1\right)\left(5k+4+1\right)\left\lbrack\left(5k+4\right)^2+1\right\rbrack\)

\(=\left(5k+3\right)\left(5k+5\right)\left(25k^2+40k+16+1\right)\)

\(=5\left(k+1\right)\left(5k+3\right)\left(25k^2+40k+17\right)\) ⋮5(13)

Từ (10),(11),(12),(13) suy ra \(q^4-1\) ⋮5

mà \(q^4-1\) ⋮3 và \(q^4-1\) ⋮8; \(q^4-1\) ⋮2

và ƯCLN(3;5;8;2)=1

nên \(q^4-1\) ⋮3*5*8*2

=>\(q^4-1\) ⋮240(14)

Từ (7),(14) suy ra \(p^4-1-\left(q^4-1\right)\) ⋮240

=>\(p^4-q^4\) ⋮240

bài này trong sách phát triển có đấy

Vì p là số nguyên tố lớn hơn 2=>p là số lẻ

=>p=2k+1

=>(p-1).(p+1)=(2k+1-1).(2k+1+1)=2k.(2k+2)=2k.2.(k+1)=4.k.(k+1)

Vì k và k+1 là 2 số tự nhiên liên tiếp.

=>k.(k+1) chia hết cho 2

=>4.k.(k+1) chia hết cho 8

=>(p-1).(p+1) chia hết cho 8(1)

Lại có: (p-1).(p+1)=p²-1

Vì p là số nguyên tố lớn hơn 3

=>p không chia hết cho 3

=>p² chia 3 dư 1

=>p²-1 chia hết cho 3

=>(p-1).(p+1) chia hết cho 3(2)

Từ (1) và (2) ta thấy:

(p-1).(p+1) chia hết cho 8 và 3

Mà (8,3)=1

=>(p-1).(p+1) chia hết cho 8.3

=>(p-1).(p+1) chia hết cho 24

Vậy (p-1).(p+1) chia hết cho 24