Three Cubes And A Square

Set $b=n$ . We then see that $a=n-1$ , and $c=n+1$ . Now, cube $n-1$ , $n$ , and $n+1$ individually:

$(n-1)^3+n^3+(n+1)^3=n^3-3n^2+3n-1+n^3+n^3+3n^2+3n+1$

$m^2=3n^3+6n$

Factorizing the above expression then yields

$m^2=3n(n^2+2)$

This expression may not seem like an improvement. However, notice that $3$ , $n$ , and $n^2+2$ are relatively prime: this implies that any solution must: $1)$ have even exponents in all prime factors $n>3$ in both $n$ and $n^2+1$ ; $2)$ have an even exponent for $n(n^2+2)$ ; and $3)$ have an odd exponent in $3$ for $n(n^2+2)$ . $2$ and $3$ follow from the fact that the prime factorization of a perfect square must be of the form

$m^2=2^{2a_1}3^{2a_2}...p_i^{2a_i}$

where $a_1, a_2....a_i$ are integers $n\geq0$ , and $2, 3...p_i$ are all of the prime factors of $m$ , trivial and non-trivial. (This results from multiplying $m=2^{a_1}3^{a_2}...p_i^{a_i}$ by itself.) However, $n$ is multiplied by $3$ and $n^2+2$ in the expression, so it follows that the exponent of $3$ in $n(n^2+2)$ is odd, and that the power of $2$ in $n(n^2+n)$ is even. The first two numbers to satisfy all three conditions are $2$ and $24$ . We have already seen $n=2$ , and $n=24$ yields

$23^3+24^3+25^3=41616=204^2$

And of course, $23+24+25=72$ .

My solution is base on solving Pell's Equation, a bit different compared to Andrei Li. (Note: All variables below are positive integers.)

Supposed the three consecutive integers are $(a-1,a,a+1)$ , and the perfect square is $n$ , so $3a(a^2+2)=n^2$

We see that $n$ must be multiple of $3$ , hence let $n=3m$ , we have $a(a^2+2)=3m^2$

Now assume $a$ is multiple of $2$ , then let $a=2b$ , $m=2k$ , thus $b(2b^2+1)=3k^2$

Now $b$ and $2b^2+1$ is coprime, we can let $\begin{cases}b=3c^2\\2b^2+1=d^2\end{cases} \text{ or }\begin{cases}b=c^2\\2b^2+1=3d^2\end{cases}$

We can find that $b=12$ is the smallest in the first system and there is no smaller solution for $b$ in second system.

Now if $a$ is odd, then we may have $\begin{cases}a=3c^2\\a^2+2=d^2\end{cases}\text{ or }\begin{cases}a=c^2\\a^2+2=3d^2\end{cases}$

We just have to check if there is any solution for $a<24$ , in fact not.

Hence, $a=24$ , then the answer is $a-1+a+a+1=3\times24=\boxed{72}$ .

Now that this is a Problem of the Week, I figured I'd repost my comment to Andrei's most-upvoted solution, since I see a lot of questions about finding all possible solutions.

TLDR: the only solutions are $[-1,0,1], [0,1,2], [1,2,3],$ and $[23,24,25]$ .

The equation $3n^3+6n = m^2,$ after the substitutions $y=3m, x=3n,$ gives $y^2 = x^3+18x.$ This is an elliptic curve with torsion subgroup ${\mathbb Z}/2$ and rank $1,$ where the nontrivial 2-torsion point is $P = (0,0)$ and a generator of the free part is $Q = (3,9).$ Now $P-Q = (6,18)$ and $P+2Q = (72,612),$ but all the other linear combinations do not give a point with integral coefficients (except for their negatives, which just correspond to flipping the sign of $y$ ). The heights of the denominators increase quite quickly, so it's pretty clear that there are no other solutions. Your favorite computer algebra system should be able to do a completely rigorous computation.

For instance, the computer algebra system MAGMA (which I used to compute the torsion subgroup and rank) has an IntegralPoints command that uses known bounds on the heights of integral points, and it verifies that the only integral points on $y^2=x^3+18x$ are $(0,0), (3, \pm 9), (6, \pm 18), (72, \pm 612).$

The sum of numbers (1, 2, ....n) squared is equal to the sum of each number cubed, or

1. $(\sum_{i=0}^{a} i)^2 = \sum_{i=0}^{a} i^3$

The left side of the equation will have numbers $1^2, 3^2, 6^2, 10^2$ and so forth.

The right side will look like this; $1^3 +2^3 +3^2 +.....+ a^3$

That's one way to look at the first two triplets given in the problem. However, any number larger than 6 in using the sum will have more than 3 cubed numbers; for example, $10^2 = 1^3 +2 ^3 +3^3 + 4^3$ . Thus, we must do something to limit the number of of cubed numbers to three. The way to do that is to subtract two squares that are 3 steps away from each other in the left side of the equation.

For example, $21^2 - 6^2 = ((1^3 +2^3 +3^3 +4^3 +5^3 + 6^3) - (1^3 +2^3 +3^3) = 4^3 +5^3 + 6^3$

The only problem now is that $21^2 - 6^2$ itself does not equal a perfect square, so it does not satisfy the premise of the problem.

To account for this, we must look at Pythagorean Triples (a, b, c) where a and b are legs and c is the hypotenuse such that c and either a or b belong to the set of (numbers)^2 from the left side of equation 1.

For example, the Pythagorean triple (6, 8, 10) contains 6 and 10, which belong to the set of (outputs)^2 from the left side of Equation 1;

$10^2 = 6^2 +8^2$ which leads to $8^2 = 10^2 - 6^2 = (1^3 +2^3 +3^3 +4^3) - (1^3 +2^3 +3^3) = 4^3$

However, (3, 4, 5) would not work here because 5 does not belong to the set (1, 3, 6, 10....), thus we cannot express 5 directly as the sum of $1^3 + 2^3...a^3$ .

So the two conditions that we must satisfy are that the Pythagorean Triple must have c and (a or b) in the set of numbers (1, 3, 6, 10....) and that c and (a or b ) must be 3 numbers apart in that set.

With these two constraints, we can then search for a triple that works, and we find that the triple (204, 253, 325) satisfies both conditions and that it is the first triple since the ones provided in the problem.

Thus, $204^2 = 325^2 - 253^2 = (1^3 + 2^3 + ... 25^2) - (1^3 + 2^3 + ... 22^2) = 23^3 + 24^3 + 25^3$

$23^3 + 24^3 + 25^3 = 204^2$ and $23 + 24 + 25 = 72$ $_\square$

This is my first time offering a solution, so please let me know how I can try and improve it, or if anything doesn't make sense. I also was not happy that I had to just sort of sift through Pythagorean triples until I found one that worked. If anyone has any ways to improve this method, please comment!

Well, I sort of cheated and ran a computer brute force simulation to find the answer. What is interesting to me is that there are only these 3 triplets till 10 Million and may be ahead.

Edit: It may take up to 10-50 Sec depending on the hardware of your PC.

The python3 code for same is:

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for i in range(0,10000000):
    x=(i**3) + (i+1)**3 + (i+2)**3
    sq=int(x**0.5)
    if (sq*sq==x):
        print(i,i+1,i+2)

This is the code I wrote in Python 3 to solve this problem:

def f(x):
    return [pow(x, 3), pow(x+1, 3), pow(x+2, 3)];

for i in range(0, 100):
    if sum(f(i))**(1/2) % 2 == 0:
        print(str([i, i+1, i+2]) + ": " + str(sum(f(i))) + " = ( " + str((sum(f(i)))**(1/2)) + ")^2").

Using this code, the next smallest set of consecutive numbers are:

[23, 24, 25]: 41616: (204.0)^2

Therefore, 23 + 24 + 25 = 72

Found the same solution as everyone here but can't figure out how to show that there is not an other one.

Check and verify,

23^3+24^3+25^3=(204)^2

Answer= (23+24+25)=72