# Show that: if $5\mid2n+1$ then $25\mid 14n^2+19n+6$

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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Try $n=3$: $5|5$ but $25\not|189$. – Hagen von Eitzen Nov 28 '12 at 19:44
@Downvoters. Please note that the OP is new - having joined last week. As such, piling on unexplained downvotes is neither constructive nor welcoming. – Bill Dubuque Nov 28 '12 at 22:20
@BillDubuque, please note the OP joined 41 days ago, not last week, and he's already asked some 6 questions or so. Besides, he was told to enhance his questions' presentation by me very few minutes after he posted it yet he chose not to. – DonAntonio Nov 28 '12 at 22:51
And something else: I usually reject substantial changes to anyone else's questions/answers as I think this is both unrespectful and incorrect, and I always advice to add comments or address the poster directly. You've changed the question directly and I think this is not the correct way to do things. – DonAntonio Nov 28 '12 at 22:54
@DonAntonio The OP has been active only a handful of days, even though their first activity was over a month ago. Further, the OP has had very little prior community feedback. So the OP is relatively new. – Bill Dubuque Nov 28 '12 at 23:35

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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I'm not seeing why that last equivalence is true, modulo $25$--though I certainly see it holding modulo $5$. – Cameron Buie Nov 28 '12 at 20:27
@Cameron Fixed, thanks. – Bill Dubuque Nov 28 '12 at 20:36
can you explain about$-11n^2+44-44 = 14n^2 +19n+6$ – geni Nov 28 '12 at 20:48
@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. – Bill Dubuque Nov 28 '12 at 21:00

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

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Or else, to avoid large numbers, $n=-2$. – André Nicolas Nov 28 '12 at 19:51
The post has been edited, $2n-1$ changed to $2n+1$. – Pedro Tamaroff Nov 28 '12 at 20:21

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.

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