Prove using congruences that $ 7\mid\left(5^{2n}+3\cdot 2^{5n-2}\right)$ , $n \ge 1$

Question

Prove using congruences that:

$$ 7\mid\left(5^{2n}+3\cdot2^{5n-2}\right)$$

(is divisible by 7)

So I'm trying to use mathematical induction to show that for all integers $n \ge 1$ but i cant prove this!!

Answer 1

Hint: Use that $$ 5^{2(n+1)}+3 2^{5(n+1)-2}=25 \cdot 5^{2n}+32 \cdot 3 \cdot 2^{5n}, $$ and take into account that $25=32 \mod 7.$

Answer 2

Claim : $$5^{2n}+3\times 2^{5n-2}\equiv 0\pmod 7$$ for $n\ge 1$

Base case : $25+3\times2^3=25+24=49\equiv 0\pmod 7$

Induction step

$$5^{2n+2}+3\times 2^{5n+3}=25\times 5^{2n}+96\times 2^{5n-2}=25\times(5^{2n}+3\times 2^{5n-2})+21\times 2^{5n-2}$$

I think you can finish the proof now.

Answer 3

mod $7$: $\quad 5^{2n}+3\cdot2^{5n-2} = 25^n +3\cdot \dfrac{32^n}{4} \equiv 4^n +3\cdot 2 \cdot 4^n \equiv 4^n - 4^n \equiv 0 $

Answer 4

Let's call the expression $f(n)$. For $n=k$, assume that $7|f(k) \implies 7|5^{2k}+3\cdot2^{5k-2}$. Then multiply the expression by $5^2\cdot2^5$ and it is still divisible by $7$. So we get $2^{5}\cdot5^{2k+2}+5^{2}\cdot3\cdot2^{5k+3}$ which is equivalent to $4\cdot5^{2k+2}+4\cdot3\cdot2^{5k+3} \pmod 7$. $7$ divides the expression and $4$ and $7$ are co-primes. So,$7|5^{2k+2}+3\cdot2^{5k+3}$ and the last expression is equal to $f(k+1) (7|f(k+1))$.