Find the smallest possible integer $n$ for which $19n+1$ and $95n+1$ are both perfect squares.
I somehow managed to show that $n$ is odd but couldn't find any solution for which both of them are perfect squares.
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$\begingroup$ Does the text imply that the integer in question be strictly positive? $\endgroup$– LucianMar 19, 2015 at 13:06
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1$\begingroup$ I ran a C program to find said number but the result was an even number... $\endgroup$– Luigi D.Mar 19, 2015 at 13:26
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$\begingroup$ It might be my error but you please post your answer. $\endgroup$– Mayank JhaMar 19, 2015 at 13:42
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1$\begingroup$ Did you try looking at the curve $5x^2-4=y^2$ and lines through the point $(1,1)$ which have rational slope? To get this, let the two squares be $x^2$ and $y^2$ and note that $95=19\cdot 5$. Since $(1,1)$ is a solution with $n=0$, it is on the curve. $\endgroup$– Michael BurrMar 19, 2015 at 14:29
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1$\begingroup$ 134232 is the smallest value $\endgroup$– aliMar 19, 2015 at 14:41
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1 Answer
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Assume,
$$95n+1 = x^2$$
$$19n+1= y^2$$
Eliminate $n$ between them and you get the Pell equation,
$$x^2-5y^2 = -4$$
I assume you know how to find all solutions $x,y$. It is then a simple matter to test them such that,
$$n = \frac{x^2-1}{95} = \frac{y^2-1}{19}$$
is an integer. The smallest ones are $n= 134232,\,920040,\,4481374227696,\, 30715795905552,\dots$ ad infinitum. (There doesn't seem to be odd n.)
Equivalently, for $m \ge 0$,
$$n = \frac{F_{18m+17}^2-1}{19} = 134232,\,4481374227696,\dots$$
$$n = \frac{F_{18m+19}^2-1}{19} = 920040,\,30715795905552,\dots$$
where $F_m$ is a Fibonacci number.
answered Mar 22, 2015 at 14:24