2025/016) For a prime $p >4$, how do you prove that $3^p-2^p \equiv 1 \pmod {42}$

 We have $ 42 = 2 * 3 * 7$

We need to show that $3^p-2^p=1   \pmod {2}\cdots(1)$

$3^p-2^p=1   \pmod {3}\cdots(2)$

and  $3^p-2^p=1   \pmod {7}\cdots(3)$ 

As $3^p$ is odd for any p and and $2^p$ us even we get

$3^p-2^p=1   \pmod {2}$

this is we have proved (1) 

 now p is odd so we have

$3^p$ is divisible by 3.

or $3^p \equiv 1 \pmod 3\cdots(4)$  

now let us look as $2^p$ we have as p is odd so p = 2k+1

$2^p = 2^{2k+1} = 4^k * 2$

so $2^p \equiv 2 \pmod 3\cdots(5)$

From (4) and (5) we have

$3^p- 2^p \equiv 1 \pmod 3$

We have proved (2)

Now we need to check for mod 7

as p is greater that 4 and a prime p is of the form 6k+1 or 6k + 5

Let us take the 2 cases 6k+ 1 and 6k-+ 5

first 6k+ 1

as 7 is co-prime to 3 and 7 we have as per Fermats Little Theorem 

$3^6 \equiv 1 \pmod 7$

so $3^{6k+1} = (3^6)^k .3 = 3 \equiv 3 \pmod 7$

 Simlilarly 

$2^6 \equiv 1 \pmod 7$

so $2^{6k+1} = (2^6)^k .3 = 2 \equiv 2 \pmod 7$

Or $3^p-2^p \equiv 1 \pmod 7$ 

So we have proved (3) for p = 6k+ 1

 Next t 6k+ 4

as 7 is co-prime to 3 and 7 we have as per Fermats Little Theorem 

$3^6 \equiv 1 \pmod 7$

so $3^{6k+5} = (3^6)^k 243 = 243 \equiv 3 \pmod 7$

 Simlilarly 

$2^6 \equiv 1 \pmod 7$

so $2^{6k+5} = (2^6)^k .32 = 32 \equiv 2 \pmod 7$

Or $3^p-2^p \equiv 221 \pmod 7$

Or   $3^p-2^p \equiv 1 \pmod 7$

So we have proved (3) for p = 6k+ 5

 We have proved (3) for both cases

as (1) (2) (3) all are proved hence Proved  

  

 

 

2025/015) Prove $4^{2n}+10n \equiv 1 \pmod {25}$

 We shall prove if by binomial expansion

we have $4^{2n}$

$= (5-1)^{2n}$

 $ = \sum_{k=0}^{2n}{2n \choose k}5^{2n-2}(-1)^{k}$

$ = \sum_{k=0}^{2n-2}{2n \choose k}5^{2n-2}(-1)^k -  {2n \choose 2n-1}*5 + 1$ separating last 2 terms

1st sum each term is divisible by $5^2$ that is 25

so we are left with 

  $4^{2n} \equiv  - 10n +1  \pmod {25}$

or  

   $4^{2n}+10n \equiv 1 \pmod {25}$