<s>[INST] Prove, for every nonnegative integer $n$, that $5^{2n} + 2^{2n} ≡ 2^{2n+1}\pmod{21}$. Is my proof correct? Thanks in advance.
Base step: n = 0
$$5^{2 \times 0} + 2^{2 \times 0} ≡ 2^{2 \times 0+1} \pmod{21}$$
$$= 1 + 1 ≡ 2 \pmod{21} \checkmark$$
Inductive hypothesis: assume $5^{2n} + 2^{2n} ≡ 2^{2n+1} \pmod{21}$ is true when $n=k$:
$$ 5^{2k} + 2^{2k} ≡ 2^{2k+1} \pmod{21}$$
Since the statement above is true we can write the following:
$$5^{2k} + 2^{2k} - 2^{2k+1} = 21x$$
Simplify and rewrite:
$$2^{2k} = 5^{2k}-21x$$
Now we proceed with the inductive step: $n = k + 1$:
$$5^{2(k+1)} + 2^{2(k+1)} ≡ 2^{2(k+1)+1} \pmod{21}$$
$$5^{2k+2} + 2^{2k+2} ≡ 2^{2k+3} \pmod{21}$$
Rephrase the statement above:
$$21 \mid (5^{2k+2} + 2^{2k+2} - 2^{2k+3})$$
$$21 \mid (5^{2k+2} - 4 \times 2^{2k})$$
Replace $2^{2k}$ with $5^{2k}-21x$:
$$21 \mid (5^{2k+2} - 4 \times (5^{2k}-21x))$$
Multiply out:
$$21 \mid (5^{2k+2} - 4\times 5^{2k} + 4\times 21x))$$
$$ = 21 \mid (25 \times 5^{2k} - 4\times 5^{2k} + 4\times 21x))$$
$$ = 21 \mid (21 \times 5^{2k} + 4\times 21x))$$
$$ = 21 \mid 21(5^{2k} + 4x))\checkmark$$
Hence
$$5^{2(k+1)} + 2^{2(k+1)} ≡ 2^{2(k+1)+1}  \pmod{21}$$
$\blacksquare$
 [/INST] The proof you gave is correct.
However, it can actually be done with a lot less work and without induction:
$$5^{2n}=\left(5^2\right)^n=25^n\equiv 4^n=2^{2n} (\mathrm{mod}\ 21)$$
Hence $5^{2n}+2^{2n}\equiv 2\cdot 2^{2n}=2^{2n+1} (\mathrm{mod}\ 21)$.
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