Can it be shown that 51 | ((10^(32n+9))-7) for all integers n>=0?

Question

This requires an application of Euler's theorem:

If n >= 1 and gcd(a,n) = 1, then a^Φ(n) ≡ 1 (mod n)

Answer 1

Yes, according Euler's Theorem and GCD(10, 3) = GCD(10, 17) = 1

10^φ(3) ≡ 10² ≡ 1 (mod 3) , hence 10¹⁶ ≡ (10²)⁸ ≡ 1 (mod 3) also, and

10^φ(17) ≡ 10¹⁶ ≡ 1 (mod 17), that means

10¹⁶  - 1 ≡ 0 (mod 3*17) or 10¹⁶ ≡ 1 (mod 51).

/The latter can be checked directly with Windows Calculator without using Euler's Theorem if You wish/

Now 10³²ⁿ⁺⁹  Ã¢Â‰Â¡ (10¹⁶)²ⁿ  * 10⁹  Ã¢Â‰Â¡ 10⁹  Ã¢Â‰Â¡ (10²)⁴ * 10  Ã¢Â‰Â¡

≡ 10(-2)⁴  Ã¢Â‰Â¡ 160 ≡ 3*51 + 7 (mod 51) or

10³²ⁿ⁺⁹ - 7 ≡ 0 (mod 51)

Answer 2

No it can't