AIME 2021 I · 第 5 题

Solution 1

Let the terms be $a-b$, $a$, and $a+b$. Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$, or $3a^2+2b^2=ab^2$. Rearranging, we get $b^2=\frac{3a^2}{a-2}$. Simplifying further, $b^2=3a+6+\frac{12}{a-2}$. Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\pm1, 2, 3, 4, 6, 12$. Looking at the first equation, we see $a>2$ since $b^2$ is positive. This means we must test $a=3, 4, 5, 6, 8, 14$. After testing these, we see that only $a=5$ and $a=14$ work which give $b=5$ and $b=7$ respectively. Thus the answer is $10+21=\boxed{031}$. ~JHawk0224

Note: You could also use synthetic division to get $b^2=\frac{3a^2}{a-2}$.

Solution 2

Let the common difference be $d$ and let the middle term be $x$. Then, we have that the sequence is

$$x-d,~x,~x+d.$$ This means that the sum of the squares of the 3 terms of the sequence is

$$(x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2.$$ We know that this must be equal to $xd^2,$ so we can write that

$$3x^2+2d^2=xd^2,$$ and it follows that

$$3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.$$ Now, we can treat $d$ as a constant and use the quadratic formula to get

$$x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.$$ We can factor pull $d^2$ out of the square root to get

$$x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.$$ Here, it is easy to figure out the values of $d$. Let $\sqrt{d^2-24} = k$, then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ note that $d$, $k$ are integers. Examining the parity, we find that $d+k$ and $d-k$ are of the same parity. Now, we solve by factoring. We can find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\sqrt{d^2-24}$ a positive integer.$^{*}$ $d=5$ gives $x=5$ and $x=\frac{10}{3}$, but we can ignore the latter. $d=7$ gives $x=14$, as well as a fraction which we can ignore.

Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=\boxed{031}$. -BorealBear, minor edit by Kinglogic

$^*$To prove this, let $\sqrt{d^2-24} = k$, then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ then remembering that $d$ and $k$ are integers see if you can figure it out. -PureSwag

Solution 3

Proceed as in solution 2, until we reach

$$3x^2+2d^2=xd^2,$$ Write

$d^2=\frac{3x^2}{x-2}$, it follows that $x-2=3k^2$ for some (positive) integer k and $k \mid x$.

Taking both sides modulo $k$, $-2 \equiv 0 \pmod{k}$, so $k \mid 2 \rightarrow k=1,2$.

When $k=1$, we have $x=5$ and $d=5$. When $k=2$, we have $x=14$ and $d=7$. Summing the two cases, we have $10+21=\boxed{031}$.

-Ross Gao

Solution 4 (Combining Solution 1 and Solution 3)

As in Solution 1, write the three integers in the sequence as $a-d$, $a$, and $a+d$.

Then the sum of the squares of the three integers is $(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2$.

Setting this equal to the middle term times the common difference squared, which is $ad^2$,

and solving for $d^2$ we get:

$3a^2+2d^2 = ad^2 \implies ad^2-2d^2 = 3a^2 \implies d^2(a-2) = 3a^2 \implies d^2 = \frac{3a^2}{a-2}$

The numerator has to be positive, so the denominator has to be positive too for the sequence

to be strictly increasing; that is, $a>2$.

For $\frac{3a^2}{a-2}$ to be a perfect square, $\frac{3}{a-2}$ must be a perfect square as well.

This means that $a-2$ is divisible by 3, and whatever left over is a perfect square.

We can express this as an equation: let the perfect square left over be $n^2$. Then:

$3n^2 = a-2$. Now when you divide the numerator and denominator by 3, you are left with

$d^2 = \frac{a^2}{n^2} \implies d = \frac{a}{n}$. Because the sequence is of integers, d must also be an

integer, which means that $n$ must divide $a$.

Taking the above equation we can solve for $a$: $3n^2 = a-2 \implies a = 3n^2+2$.

This means that $3n^2+2$ is divisible by $n$. $3n^2$ is automatically divisible by $n$, so

$2$ must be divisible by $n$. Then $n$ must be either of $\{1,2\}$. Plugging back into the equation,

$n = 1 \implies a = 5 \implies d = 5$, so $a+d = 5+5 = 10$.

$n = 2 \implies a = 14 \implies d = 7$, so $a+d = 14+7 = 21$.

Finally, $10+21 = \boxed{031}$

-KingRavi

Solution 5

Following from previous solutions, we derive $3x^2+2a^2=xa^2.$ We divide both sides to get $3\left(\frac{x}{a}\right)^2+2=x.$ Since $x$ is an integer, $\frac{x}{a}$ must also be an integer, so we have $x=pa$, for some factor $p$. We then get $3p^2+2=pa.$ We then take this to modulo $p$, getting $2\equiv 0 \pmod p.$ The only possibilities for $p$ are therefore 1 and 2. We plug these into $3p^2+2=pa$, for $a=5$ and $x=5$, giving us the sequence $0,5,10$, or $2a=14$ and $x=14$, for the sequence $7,14,21.$ $10+21 = \boxed{031}.$

-RYang2

Video Solution 1

https://youtu.be/92dvTKV1nPc

~MathProblemSolvingSkills.com

Video Solution 2

https://www.youtube.com/watch?v=I43RH5DUa1I

Video Solution 3

https://youtu.be/M3DsERqhiDk?t=1465