Разность 97^2025-5^2025 делится на 91. Докажите это, не пиша

[Гость]

Разность 97⁽²⁰²⁵⁾ − 5⁽²⁰²⁵⁾ делится на 91. Докажите это, не пиша компьютерной программы и не пользуясь катькулятором.

[Гость]

Мы будем работать над формулами :
[tex]x^{5 }- у^{5 }= (х-у)( x^{4 }+ x^{3 }у+ x^{2 } у^{2 } +ху^{3 } +у^{4 })[/tex]

[tex]a^{3 } -b^{3 } =(a-b)(a^{2 }+ab+ b^{2 })[/tex]
___________________________________________
[tex]( 97^{405 } )^{5 }- ( 5^{405 } )^{5 }[/tex]=

=[tex][97^{405 } -5^{405 }]( 97^{405.4 } +97^{405.3 } .5^{405 } +97^{405.2 } .5^{405.2 } +97^{405 } .5^{405.3 } +5^{405.4 })[/tex]=
Пусть второй множитель будет А .

=[ [tex]( 97^{81 } )^{5 } - ( 5^{81 } )^{5 } ]A=[/tex]

=[ [tex]97^{81 } -5^{81 } ].BA= [ ( 97^{27 } )^{3 } -( 5^{27 } )^{3 } ].BA=[/tex]

=[tex][97^{27 } -5^{27 } ]( 97^{27.2 } +97^{27 } .5^{27 } +5^{27.2 }).BA=[/tex]

=[ [tex]( 97^{9 } )^{3 } - ( 5^{9 } )^{3 } ].CBA=[/tex] [tex][ 97^{9 } -5^{9 } ]ECBA[/tex]=

=[tex][ ( 97^{3 } )^{3 } -( 5^{3 } )^{3 } ]ECBA=[/tex]

=[tex][ 97^{3 } -5^{3 } ]( 97^{3.2 } +97^{3 } .5^{3 } +5^{3.2 } )ECBA=[/tex]

=[tex][ 97^{3 } -5^{3 } ]FECBA=[/tex]

=[ 912 673-125 ]FECBA= 912 548.FECBA=

=91 .10 028.FECBA

Мысли из Болгарии .

[Гость]

Key facts:

- 91 = 7 × 13.
- To show 97^2025 − 5^2025 is divisible by 91, it’s enough to show it’s divisible by both 7 and 13.

Step 1: Mod 7

- 97 ≡ 97 − 98 = −1 (mod 7), so 97 ≡ −1 (mod 7).
- 5 ≡ 5 (mod 7).
- Then 97^2025 ≡ (−1)^2025 = −1 (mod 7).
- And 5^2025 ≡ 5^odd ≡ 5 (mod 7).
- So 97^2025 − 5^2025 ≡ (−1) − 5 = −6 ≡ 1 (mod 7)? Wait—check 5 (mod 7) powers: better use 5 ≡ −2 (mod 7).- 5^2025 ≡ (−2)^2025 = −2 (mod 7).
- Then 97^2025 − 5^2025 ≡ (−1) − (−2) = 1 (mod 7). That suggests not divisible by 7—so we need to correct.

Let’s compute carefully:

- The key is to compare 97 and 5 modulo 7:- 97 ≡ −1 (mod 7).
- 5 ≡ 5 ≡ −2 (mod 7).


- For odd exponent n, (−1)^n = −1 and (−2)^n = −2.
- Then difference is (−1) − (−2) = 1 (mod 7), which is not 0.

This means we should not compare directly; instead use the factorization trick:

- 97 ≡ −1 (mod 7) and 5 ≡ −2 (mod 7). Note that −1 ≡ 6 and −2 ≡ 5 mod 7.
- Observe that 97 ≡ 5 + 1 (mod 7), since 6 ≡ 5 + 1 (mod 7).
- Use the binomial factor: a^n − b^n is divisible by a − b when n is odd.
- Specifically, if n is odd, then a − b divides a^n − b^n.

Take a = 97 and b = 5:

- Since 2025 is odd, 97 − 5 = 92 divides 97^2025 − 5^2025.
- But we need divisibility by 91, not 92, so that’s not directly helpful.

Let’s use modulo 13 and 7 separately via Euler/Fermat or orders.

Better approach: Reduce bases modulo 91 by Chinese remainder decomposition.

Mod 13:

- 97 ≡ 97 − 91 = 6 (mod 13).
- 5 ≡ 5 (mod 13).
- So we want 6^2025 − 5^2025 mod 13.
- Note 6 ≡ −7 ≡ −7 (not helpful). Instead use orders:- φ(13) = 12. For any a coprime to 13, a^12 ≡ 1 (mod 13).
- 2025 mod 12: 2025 = 12×168 + 9, so 2025 ≡ 9 (mod 12).
- Compute 6^9 − 5^9 mod 13.
- Use small reductions: 6^2 = 36 ≡ 10 (mod 13), 6^4 ≡ 10^2 = 100 ≡ 9 (mod 13), 6^8 ≡ 9^2 = 81 ≡ 3 (mod 13), so 6^9 ≡ 3×6 = 18 ≡ 5 (mod 13).
- For 5: 5^2 = 25 ≡ 12 (mod 13) ≡ −1, so 5^4 ≡ 1, hence 5^8 ≡ 1, thus 5^9 ≡ 5 (mod 13).
- Therefore 6^2025 − 5^2025 ≡ 5 − 5 ≡ 0 (mod 13). Pythagoras theorem

Mod 7:

- 97 ≡ −1 (mod 7), 5 ≡ −2 (mod 7).
- Again use order arguments:- For any a coprime to 7, a^6 ≡ 1 (mod 7).
- 2025 mod 6: 2025 = 6×337 + 3, so 2025 ≡ 3 (mod 6).
- Compute (−1)^3 − (−2)^3 = (−1) − (−8) = −1 + 8 = 7 ≡ 0 (mod 7).


- Thus 97^2025 − 5^2025 ≡ 0 (mod 7).

Since the difference is 0 modulo both 7 and 13, it is divisible by 7×13 = 91.

Summary:

- Mod 13: Reduce exponents using Fermat’s theorem to show 6^2025 ≡ 5^2025 (mod 13).
- Mod 7: Reduce exponents mod 6 to show (−1)^2025 − (−2)^2025 ≡ 0 (mod 7).
- Therefore divisible by 91.