perplexus.info :: Numbers : Powers of 2 vs some factorials

Powers of 2 vs some factorials (Posted on 2022-06-08)

Prove that equation
2^x-y! = 2
Has only 2 valid integer solutions.

Bonus: How about x²-y! = 2 ?

See The Solution Submitted by Ady TZIDON

Rating: 3.5000 (2 votes)

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D1 solution Comment 5 of 5 |

2^2-2!=4-2=2
2^3-3!=8-6=2

For all y>=4, y! is divisible by 4. Therefore, y!+2 cannot be a power of 2.

2^2-2!=4-2=2
-2^2-2!=4-2=2

For all y>=4, y! is divisible by 4. There is no x such that x^2=2 mod 4. Therefore, x^2-y! cannot be 2.

Posted by Math Man on 2022-06-08 12:23:37

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