How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?
Answer
You can prove this by induction.
First, show that this is true for $n=1$:
$5^{2\cdot1+1}+3^{1+2}\cdot2^{1-1}=19\cdot8$
Second, assume that this is true for $n$:
$5^{2n+1}+3^{n+2}\cdot2^{n-1}=19k$
Third, prove that this is true for $n+1$:
$5^{2(n+1)+1}+3^{n+1+2}\cdot2^{n+1-1}=$
$5^{2+2n+1}+3^{1+n+2}\cdot2^{1+n-1}=$
$5^{2}\cdot5^{2n+1}+3^{1}\cdot3^{n+2}\cdot2^{1}\cdot2^{n-1}=$
$5^{2}\cdot5^{2n+1}+3^{1}\cdot2^{1}\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot5^{2n+1}+\color\green{6}\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot5^{2n+1}+(\color\green{25-19})\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot5^{2n+1}+25\cdot3^{n+2}\cdot2^{n-1}-19\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot(\color\red{5^{2n+1}+3^{n+2}\cdot2^{n-1}})-19\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot\color\red{19k}-19\cdot3^{n+2}\cdot2^{n-1}=$
$19\cdot25k-19\cdot3^{n+2}\cdot2^{n-1}=$
$19\cdot(25k-3^{n+2}\cdot2^{n-1})$
Please note that the assumption is used only in the part marked red.
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