Minimum Common Difference

Say the three numbers are $a,a+d^2,a+2d^2$

with both $a,d$ positive integers.

Then the question asks for the smallest value of $d^2$ (the common difference) such that $\left(a+2d^2\right)^2-\left(a+d^2\right)^2=a^3$

Tidying the left hand side, $d^2 \left(2a+3d^2\right)=a^3$

Now, let $a=kd$ (note that $k$ may not be an integer, but is certainly rational).

The equation becomes $d^2 \left(2kd+3d^2\right)=k^3 d^3$

Cancelling, $k^3-2k-3d=0$

Now, this is a monic polynomial in $k$ with integer coefficients. By the integral root theorem, $k$ is either irrational or an integer. We know it's not irrational, by definition, so it must be an integer.

Hence $d=\frac{k^3-2k}{3}$ for some integer $k$ .

Since $k^3-2k=k(k^2-2)$ and $k^2-2 \not\equiv 0 \pmod 3$ , we need $k$ to be a multiple of $3$ ; take $k=3t$ .

So all solutions have the form $d=9t^3-2t$ , $a=3td$ .

The smallest of these comes from $t=1$ , giving $d=7$ , $a=21$ . The three terms of the sequence are $21,70,119$ with a common difference of $\boxed{49}$ .

Let the smallest number in the arithmetic progression be $a$ and the common difference be $d$ . Then we have:

$\begin{aligned} (a+2d)^2 - (a+d)^2 & = a^3 \\ (a+2d+a+d)(a+2d-a-d) & = a^3 \\ (2a+3d)d & = a^3 \\ 3d^2 + 2ad - a^3 & = 0 \end{aligned}$

$\begin{aligned} \implies d & = \frac {-2a + \sqrt{4a^2+12a^3}}6 = \frac {a(\sqrt{1+3a}-1)}3 \end{aligned}$

For $d$ to be an integer, $1+3a$ must be a perfect square. Let $b^2 = 1+3a$ . Then we have:

$\begin{aligned} d & = \frac {a(b-1)}3 & \small \blue{\text{Note that }a = \frac {b^2-1}3} \\ & = \frac {(b^2-1)(b-1)}9 \\ & = \frac {(b-1)^2(b+1)}9 \end{aligned}$

Now we note that $d$ is a perfect square when $\dfrac {b+1}9$ is a perfect square. Bonus: That is $d$ is a perfect square whenever $b+1 = 9n^2$ , where $n$ is a positive integer. And the smallest perfect square $d$ is when $n=1$ or $b=8$ , $\implies d = \dfrac {(8-1)^2\cancel{(8+1)}}{\cancel 9} = 7^2 = \boxed{49}$ .

Let the sequence be $a, a+d^2, a + 2d^2$

$\begin{aligned} (a+2d^2)^2 - (a+d^2)^2 &= a^3 \\ (2a+3d^2) (d^2) &= a^3 \\ 3(d^2)^2 + 2a (d^2) - a^3 &= 0\\ d^2 &= \cfrac{-2a + \sqrt{ 4a^2 + 12 a^3}}{6} \hspace{5mm} \color{#3D99F6}{\text{neglect -ve } }\\ &=\cfrac{-2a + 2a\sqrt{ 1 + 3 a}}{6} \\ &= \cfrac{-a + a\sqrt{ 1 + 3 a}}{3} \\ \text{ as }\textcolor{#3D99F6}{ (3q+2)^2 = 3Q + 1 } &\text{ for some integer q, Q, the numerator will be } -a + a(3Q+2) = (3Q+1) a \rightarrow \boxed{a=3k} \\ &= \cfrac{-(3k) + (3k)\sqrt{ 1 + 3(3k)}}{3} \\ &=-k+k\sqrt{1+9k}\\ \text{ as }\textcolor{#3D99F6}{ (9m \pm 1)^2 = 9k + 1 } &\text{ for some integer m, } \sqrt{1+9k} = (9m \pm 1) \hspace{5mm} \text{[1]}\\ \therefore d^2 &= \left\{\begin{array}{c}&9mk \\ & (9m-2)k \end{array} \right. \\ \text{ By [1] } (9m \pm 1)^2 &= 9k + 1 \implies 9k =(9m \pm 1)^2 - 1 =\left\{\begin{array}{c}&(9m+2)(9m) \\ &(9m) (9m-2) \\ \end{array} \right. \\ \implies k &= \left\{\begin{array}{c}&(9m+2)(m) \\ &(9m-2)(m) \\ \end{array} \right. \\ \implies d^2 &= \left\{\begin{array}{c}&9m^2(9m+2) = 3^2 m^2 (9m+2) \hspace{5mm} \text{case 1} \\ &(9m-2)^2(m) = (9m-2)^2 (m) \hspace{5mm} \text{case 2} \\ \end{array} \right. \\ \end{aligned}$

For case 1, we need $9m+2 = x^2$ for some integer $x,$

there is no solution as $(3m)^2 = 9m^2, (3m\pm 1)^2 = 9m^2 \pm 6m+1 \Rightarrow 6m+1=9Q+2 \Rightarrow 6m = 9Q + 1 \Rightarrow 0 \equiv 1 \pmod 3 !$

For case 2, we need $m = x^2$ , there are infinite many solutions.

$\therefore d^2 = (9x^2-2)^2 (x^2)$ for integer $x, a = 3k = 3 (9x^2-2)(x^2) \\ \min. d^2 = (9-2)^2 (1) = 7^2 = \boxed{49}$ while $a = 3(9-2)(1) = 21$

The next $d^2 = (9(2)^2-2)^2 (2^2) = 34^2 \cdot 2^2 = 68^2 = 4624, a=3(9\times4-2)\times4 = 408$

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$9k+1 = (3m\pm 1)^2 \implies 9k+1 = 9m^2 \pm 6m + 1 \implies 9k = 9m^2 \pm 6m \implies m = 3M, \therefore \text{ set } 9k+1 = (9m\pm 1)^2$