Show that: if $ 5\mid(2n+1),\; $ then $25\mid (14n^2+19n+6) $.
[Note: $\ $ was $\,\ 5\mid (2n\color{#C00}- 1)\,\ldots$ in original version. Some answers and comments below give counterexamples to the original version. --moderator]
Show that: if $ 5\mid(2n+1),\; $ then $25\mid (14n^2+19n+6) $.
[Note: $\ $ was $\,\ 5\mid (2n\color{#C00}- 1)\,\ldots$ in original version. Some answers and comments below give counterexamples to the original version. --moderator]
Write $14n^2+19n+6 = (2n+1)^2 + 5(2n+1)(n+1)$ and note that each term on the right hand side is divisible by 25.
this doesn't hold. counterexample: $n=3 \implies 14 n ^2 + 19 n + 6 = 189$.
Hint $\rm\,\ 14n^2\!+\!19n\!+\!6 = ({2n\!+\!1})(7n\!+\!6)\ $ and $\rm\:5\mid 2n\!+\!1\:\Rightarrow\:5\mid 7n\!+\!6 = 2n\!+\!1\! + 5(n\!+\!1)\ \ $ QED
It's a special case $\rm\:p=5,\ a,b = 7n\!+\!6,\, 2n\!+\!1\:$ of this
Lemma $\ $ If prime $\rm\:p\mid a\!-\!b\:$ then $\rm\:p\mid ab\:\Rightarrow\: p^2\mid ab.$
Proof $\rm\,\ p\mid a\!-\!b\:$ implies $\rm\:p\mid a\iff p\mid b,\:$ so $\rm\:p\mid a,b\:$ (else $\rm\:p\nmid a,b\:\Rightarrow\:p\nmid ab\:$ by $\rm\:p\:$ prime).
OR $ $ if you know mod arithmetic: if $\rm\ 5\mid 2n\!+\!1\:$ then, mod $5\!:\,$ $\rm\:2n\equiv -1\equiv 4\,$ so $\rm\:n\equiv 2,\:$ thus
$\rm 5\mid n\!-\!2\ \Rightarrow\ 5^2\mid 14(n\!-\!2)^2 \equiv 14n^2\!+19n\!+\!6\!\!\pmod{25}$