I have tried doing:
$$n+5=d*x$$ $$n^2+7=d*y$$ for x and y when they are both integers. $$\frac{n+5}{x}=\frac{n^2+7}{y}$$ $$y(n+5)=x(n^2+7)$$
I am stuck here, does anyone know how to continue the proofing? Thanks
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For such problems much of the time all you need to know is that $d$ will divide multiples and sums of whatever you know $d$ already divides. – Derek Luna Jul 13 '19 at 05:43
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https://math.stackexchange.com/questions/3289844/if-i-know-that-b-is-a-multiple-of-32-then-how-can-i-show-that-b5-and-b2/3289846?noredirect=1#comment6766596_3289846 – nonuser Jul 13 '19 at 05:54
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How is this a duplicate of that post? – Derek Luna Jul 13 '19 at 05:57
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Strange people @DerekLuna – nonuser Jul 13 '19 at 07:15
4 Answers
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$d|n^2+7-(n+5)(n-5)$, where the right hand side is 32.
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A quick solution...
We note that $d|(n+5)\implies d|(n+5)^2\implies d|(n^2+10n+25)$, but $d|(n^2+7)$ and hence we get,$$d|(n^2+10n+25-(n^2+7))\implies d|(10n+18)$$Also, $d|(n+5)\implies d|(10n+50)$, but $d|(10n+18)$ and hence we get,$$d|(10n+50-(10n+18))\implies d|32$$This completes the proof...
Anand
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Yes, you are right, but the above technique always works. Just try to eliminate $n$... – Anand Jul 13 '19 at 05:52
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We know that: $$d | n + 5, \quad d | n^2 + 7,\quad \Rightarrow d | (n+5)^2 - (n^2+7) = 10(n+5) - 32$$ This means that: $$d | 10(n+5) - 32,\quad \Rightarrow d | 32$$ beacause $d | n+5$.
Lisbeth
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Not me, but I am guessing it is because people are giving slightly different and longer ways to do the problem when there is the perfectly good answer by user negative one over twelve. – Derek Luna Jul 13 '19 at 06:11
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Well, if $d|n+5$ and $d|n^2 +7$ then $d|j(n+5) + k(n^2 + 7)$ for all integers $j,k$. So if we can find integers $j,k$ so that $j(n+5) + k(n^2 +7) = 32$ we'd be done.
We want to have the $n$ and $n^2$ terms cancel out and we can start to do that by considering multiplying then $n+5$ by $n$ and subtracting the $n^2 + 7$ term.
That is $n(n+5) - (n^2+7) = (n^2 + 5n) - (n^2 +7) = 5n+7$.
To get rid of the $5n$ we can subtract multiple $n+5$ by $-5$ so we can get:
$n(n+5) - (n^2 + 7) - 5(n+5) = (5n+7) - (5n+25)=-32$. And to get the final result of $32$ we simply multiply the whole thing by $-1$.
$32 = -n(n+5) +(n^2 + 7) + 5(n+5) =(5-n)(n+5) + (n^2+7)$.
So if $j = (5-n)$ and $k = 1$ then
$d|j(n+5) + k(n^2 + 7) = (5-n)(n+5) + (n^2+7) = (25 - n^2) + (n^2 + 7) = 32$.
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To continue what you were doing:
$n + 5 = x\times d$ and $n^2 + 7 = y\times d$
So $-n^2 + 25 = (5-n)(n+5) = (5-n)xd$.
So $32 = (-n^2 + 25) + (n^2 + 7) = (5-n)xd + yd = [(5-n)x + y]d$.
