Prove using congruences that $ 7\mid\left(5^{2n}+3\cdot 2^{5n-2}\right)$ , $n \ge 1$ 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!!
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.
Let's call the expression f(n). For n=k, assume that 7|f(k) --> 7|5^(2k)+3*2^(5k-2). Then multiply the expression by $5^2*2^5$ and it is still divisible by 7. So we get 2^5*5^(2k+2)+5^2*3*2^(5k+3) which is equivalent to 4*5^(2k+2)+4*3*2^(5k+3) mod 7. 7 divides the expression and 4 and 7 are co-primes. So,7|5^(2k+2)+3*2^(5k+3) and the last expression is equal to f(k+1) (7|f(k+1)).