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Suppose $m$ is an integer such that $9m^2 + 25m + 26$ is the product of two consecutive integers. Find $m.$

I first let $k$ be equal to the larger of the two consecutive integers so that I can set up the equation $9m^2 + 25m + 26 = k(k-1).$ However, I am unsure where to go from here. Can someone give me a hint please?

questionasker
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    A common method in this type of problems is to find bounds. For example if $k<3m$ then $k(k-1) <3m(3m-1) <9m^2+25m+26$. Likewise, if $k>3m+6$ then $k(k-1)>9m^2+25m+26$ so $k \in {3m,3m+1,3m+2,3m+3,3m+4,3m+5,3m+6}$. Give it a shot this way with tighter bounds possibly. – cgss Oct 23 '20 at 17:16
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    I know it may sound lame but sometimes in various number-theory questions we check for small solutions, and because this question tells you to find only one m such that it is the product of 2 consecutive integers, I would start by plugging $m = 0, \pm 1, \pm 2 $ This is not a general way to solve it, but because this question is not general - I don't think it is that wrong – CSch of x Oct 23 '20 at 17:20
  • I question the interpretation of the previous comment. The problem statements "Suppose $m$ is..." and "Find $m$" might reasonably be interpreted as "Find all $m$". – user2661923 Oct 23 '20 at 17:39
  • I'm looking at WA right now, and there seems to be more than one solution for $m.$ However, I was thinking along the lines of applying the discriminant to the do the problem. – questionasker Oct 23 '20 at 17:41

2 Answers2

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The key insight is completing the squares.

Let $y=18m+25$. Then $9m^2 + 25m + 26 = k(k-1)$ becomes $y^2+311=(6k-3)^2-9$. So, it boils down to writing $320$ as difference of two squares, which is easy given the factorization of $320$. There are only seven solutions of $320=ab$ with $0\le a\le b$. Only one works and gives $m=3$ and $k=14$. If you allow negative numbers, then there is also $m=3$ and $k=-13$, $m=-2$ and $k=-3$, $m=-2$ and $k=4$.

lhf
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One possible way :

Let $M=9m^2 + 25m + 26$

Then discriminant, $D$ of $k^2 \pm k - M=0$ is a perfect square for integer root(s) $k$. So $D = 1 + 4M$ is a perfect square.

$$4M + 1 = 36m^2 + 100m + 105$$

Compare with nearby squares $$ \begin{align} (6m + 8)^2 \cdots 36m^2 + 100m + 105 \cdots (6m + 9)^2 \\ 36m^2 + 96m + 64 < 36m^2 + 100m + 105 \cdots 36m^2 + 108m + 81 \\ \end{align} $$

For $$ 36m^2 + 100m + 105 = 36m^2 + 108m + 81 $$

$m=3$ and $k=14$, $-13$.

cosmo5
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  • @user2661923, you're right. Manual check can't be avoided. So $m=-2$ is also acceptable answer, giving $k=4, -3$. – cosmo5 Oct 23 '20 at 20:25