Given prime number $p$ and $n \in\mathbb{N}^*$ such that $n>p$, such that:$$(3\sqrt{p+n}+3p^2-n)(\frac{3n+1}{p}+\frac{1}{n})=3p(3n+1)+8(\frac{n}{p}+1)$$ identify the numbers $n$ and $p$. Here's how far I have come so far with this problem. First, I got rid of the paranteses: $$\frac{3(3n+1)\sqrt{p+n}}{p}+3p(3n+1)-\frac{n(3n+1)}{p}+\frac{3\sqrt{n+p}}{n}+\frac{3p^2}{n}-1=3p(3n+1)+8\frac{n}{p}+8,$$ or: $$\frac{3(3n+1)\sqrt{p+n}}{p}-\frac{n(3n+1)}{p}+\frac{3\sqrt{n+p}}{n}+\frac{3p^2}{n}-1=8\frac{n}{p}+8/*np$$ Moving everything to the LHS, and multiplying by $np$ we get:$$3n(3n+1)\sqrt{n+p}-n^2(3n+1)+3p\sqrt{n+p}+3p^3-8n^2-9np=0,$$ or: $$3n(3n+1)\sqrt{n+p}-3n^3+3p^3-9n^2-9np+3p\sqrt{n+p}=0/:3$$ $$n(3n+1)\sqrt{n+p}-n^3+p^3-3n^2-3np+p\sqrt{n+p}=0$$ After factoring, we get: $$\sqrt{n+p}(3n^2+n+p)+(p-n)(p^2+np+n^2)-3n(n+p)=0$$ And I got stuck. Squaring has led me nowhere at all. I know that in order to have solutions $n+p$ has to be a perfect square. With this I've tried to observe some things, but with no succes.
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$\begingroup$ You are right. My bad. $\endgroup$– fikoooApr 6 at 12:11
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1 Answer
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Let $\sqrt{n+p} =k$ then
$$k(3n^2+k^2)+(p-n)(p^2+np+n^2)-3nk^2=0$$ so
$$3kn(n-k)+ k^3+p^3-n^3=0$$
and thus $$p^3 = (n-k)^3\implies p= n-k$$
This should be easy now...
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$\begingroup$ I expect I’m missing the obvious, but I can see how there are any solutions due to $n>p$ $\endgroup$ Apr 6 at 15:11
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$\begingroup$ This criteria comes from the relation between $p$ and $n$ in my opinion. Substituting back we get that $n=\frac{k²+k}{2}$ and $p=\frac{k²-k}{2}$. Since k is a natural number, it is not that hard to show that $n>p$ $\endgroup$– fikoooApr 6 at 16:27
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$\begingroup$ @fikooo But a number of the form $\tfrac{k^2-k}{2}$ is obviously never prime for $k>3$... $\endgroup$– ServaesApr 6 at 21:31
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$\begingroup$ It is true, but the problem asks for particular $p$ prime numbers as a solution as well as $n$ natural numbers. I do not think that having only {3} for $p$ would be a problem whatsoever. $\endgroup$– fikoooApr 7 at 4:43