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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]

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    @DonAntonio You have the facts wrong. The question was corrected by the OP, not by I (though I did infer the correction in the first version of *my* answer). My note in the question merely clarifies the OP's too-terse remark about the prior typo (it said only "EDIT 2n+1 is true"). Without such, many reader's would probably be quite confused by the comments and answers to the original version (with the typo).2012-11-28

3 Answers 3

1

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.

3

this doesn't hold. counterexample: $n=3 \implies 14 n ^2 + 19 n + 6 = 189$.

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    The post has been edited, $2n-1$ changed to $2n+1$.2012-11-28
2

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

$\rm 5\mid n\!-\!2\ \Rightarrow\ 5^2\mid 14(n\!-\!2)^2 \equiv 14n^2\!+19n\!+\!6\!\!\pmod{25}$

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    @ing The expressions are congruent mod $25$ since e.g. $\rm\,-11n^2 \equiv 14n^2\:$ since $\, -11\equiv 14.\:$ Therefore $\rm\,25$ divides LHS $\rm\:\Rightarrow\:0\equiv LHS\equiv RHS\ (mod\ 25),\:$ so $\,25\,$ divides both. Similarly for the expressions in my latest edit.2012-11-28