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- Thread starter MarkFL
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- Feb 7, 2012

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Partial solution:It can be shown that when $5n+1$ and $7n+1$ (where $n\in\mathbb{N}$) are both perfect squares, then $n$ is divisible by $24$.

Find a method for generating all such $n$.

The numbers $p_k$ satisfy the recurrence relation $p_k = 12p_{k-1} - p_{k-2}$, and $q_k$ satisfies the same recurrence relation. That enables you to generate all the values of $n$, but I have not been able to find an explicit formula for $n_k.$

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Good work,Partial solution:

The numbers $p_k$ satisfy the recurrence relation $p_k = 12p_{k-1} - p_{k-2}$, and $q_k$ satisfies the same recurrence relation. That enables you to generate all the values of $n$, but I have not been able to find an explicit formula for $n_k.$

I was not necessarily asking for a closed form solution; a recursive algorithm fits the bill for finding a method to generate the sequence.

I will post my solution within 24 hours.

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(1) \(\displaystyle 5n+1=p^2\)

Since both squares have the same parity, their difference is even, and so we may let:

(2) \(\displaystyle 7n+1=(p+2k)^2\) where \(\displaystyle k\in\mathbb{N}\)

Subtracting (1) from (2), we obtain:

\(\displaystyle 2n=(p+2k+p)(p+2k-p)=4k(p+k)\)

\(\displaystyle n=2k(p+k)\)

Substituting for $n$ into (1), we have:

\(\displaystyle 5(2k(p+k))+1=p^2\)

\(\displaystyle p^2-10kp-\left(10k^2+1 \right)=0\)

Taking the positive root for $p$, we obtain:

\(\displaystyle p=5k+\sqrt{35k^2+1}\)

A sequence defined recursively by:

\(\displaystyle A_{n+1}=sA_{n}+tA_{n-1}\)

will have limiting values of:

\(\displaystyle L=\lim_{n\to\pm\infty}\frac{A_{n+1}}{A_{n}}=\frac{s\pm\sqrt{s^2+4t}}{2}\)

Observing that for \(\displaystyle s=12,t=-1\) we have:

\(\displaystyle L=\lim_{n\to\pm\infty}\frac{A_{n+1}}{A_{n}}=6\pm \sqrt{35}\)

Thus, we find $p$ may be given in closed form by:

\(\displaystyle p_{m}=c_1\left(6+\sqrt{35} \right)^m+c_2\left(6-\sqrt{35} \right)^m\) where $p_0=1,\,p_1=11$

The initial values are obtained from $n=0,\,24$. We may now determine the parameters $c_i$ from the initial values:

\(\displaystyle p_{0}=c_1+c_2=1\,\therefore\,c_2=1-c_1\)

\(\displaystyle p_{1}=c_1\left(6+\sqrt{35} \right)+\left(1-c_1 \right)\left(6-\sqrt{35} \right)=11\)

Hence:

\(\displaystyle c_1=\frac{7+\sqrt{35}}{14},\,c_2=\frac{7-\sqrt{35}}{14}\)

And so we have:

\(\displaystyle p_{m}=\frac{1}{14}\left(\left(7+\sqrt{35} \right)\left(6+\sqrt{35} \right)^m+\left(7-\sqrt{35} \right)\left(6-\sqrt{35} \right)^m \right)\)

Now, using the relationship between $n$ and $p$ in (1), we have:

\(\displaystyle n=\frac{p^2-1}{5}\)

Hence:

\(\displaystyle n_m=\frac{\left(\frac{1}{14}\left(\left(7+\sqrt{35} \right)\left(6+\sqrt{35} \right)^m+\left(7-\sqrt{35} \right)\left(6-\sqrt{35} \right)^m \right) \right)^2-1}{5}\)

- Oct 16, 2012

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Sorry for interrupting with a somewhat "useless" post. Carry on.

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Difference equations...they're comin' for ya...

Sorry for interrupting with a somewhat "useless" post. Carry on.

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- Mar 5, 2012

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Weren't you posting problems with differential equations?

Sorry for interrupting with a somewhat "useless" post. Carry on.

I found this post by

Sorry for another interrupting an somewhat useless post.