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For how many integer values of $n$ , $n^2 -5n-3$ is a square number I don't know what to do

Parcly Taxel
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John
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  • Similar to https://math.stackexchange.com/questions/2272982/if-n24n10-is-a-perfect-square-then-find-the-possible-integer-values-of-n/2272991. – lhf May 12 '17 at 13:28
  • See: https://math.stackexchange.com/questions/524365/how-many-integer-values-of-n-are-possible-for-n225n19-to-be-a-perfect-squ – lab bhattacharjee May 12 '17 at 14:02

2 Answers2

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Hint: If $n^2 -5n-3=m^2$, complete the square after multiplying both sides by $4$.

Solution:

We get $37=(2n-5-2m)(2n-5+2m)$. Since $37$ is prime, the only possible factors are $\pm 1, \pm37$, and there are at most $4$ solutions, which can be found by solving a linear system.

lhf
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As an alternative to lhf's answer, if $n^2-5n-3=m^2$ (with $m\ge0$), then

$$n={5\pm\sqrt{5^2+4(m^2+3)}\over2}={5\pm\sqrt{4m^2+37}\over2}$$

which is an integer if and only if $4m^2+37$ is a square. But $4m^2+37=(2m)^2+37$ can be a square only if it's at least as big as the next square after $(2m)^2$, that is, only if

$$4m^2+37\ge(2m+1)^2=4m^2+4m+1$$

This limits the options for $m$ to $0\le m\le9$. So we have at worst $10$ possibilities to check.

We can eliminate the even numbers $m\in\{0,2,4,6,8\}$ because if $m$ is even then $4m^2+37\equiv5\not\equiv1$ mod $8$ (that is, odd squares are all congruent to $1$ mod $8$). This leaves $m\in\{1,3,5,7,9\}$, for which $\sqrt{4m^2+37}$ can be checked one by one.

If you like, you can trim the set to check to $m\in\{3,9\}$ since $4m^2+37\equiv2$ mod $3$ if $3\not\mid m$. At this point it's easy enough to check that $4\cdot3^2+37=73$ is not a square, leaving only $m=9$ to consider....

For the final answer (namely the number of integer values of $n$), keep in mind that the quadratic formula gives two values of $n$ for each value of $m$.

Barry Cipra
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