6
$\begingroup$

I (non-mathematician) asked a similar kind of question 5 days ago. Now I revisit the case in a different manner. The powerful numbers may be written in the form $A^2B^3$, where $A$ and $B$ are positive integers. Erdös conjectured in 1975 that there do not exist three consecutive powerful integers. The problem is notoriously tough. My simple calculations seem to prove the conjecture, and hence I would like someone to try to find the mistake (I failed in my search), which is probably more probable than not. And please point out where you found the mistake. Still, I hope my calculations are entertaining (at least they should be easy to follow).

Two powerful numbers, a difference of which is $2$, may be written as follows:

$$(A + N_2)^2 (B + M_2)^3 - A^2 B^3 = 2 \tag1$$

where $N_2$ and $M_2$ are integers, and they must be even because the largest powerful number is odd [Beckon]. We can now perform the following replacements $N_2=2N$ and $M_2=2M$.

$$(A + 2N)^2 (B + 2M)^3 - A^2 B^3 = 2 \tag2$$

We may expand the terms and divide both sides of equation $(2)$ by $2$.

$$3MA^2 B^2 + 2A^2M(3 B M + 2 M^2) + 2 A N (B + 2 M)^3 + 2 N^2 (B + 2 M)^3=1 \tag3$$

In equation $(3)$, the only possible term that could be odd is $3MA^2B^2$. Hence, $M$ must be odd. Two powerful numbers, a difference between which is $2$, may also be written as follows:

$$(A^2+n_2)(B^3+m_2)-A^2B^3=2 \tag4$$

Expanding the terms of equation $(4)$ and dividing both sides by $2$, we get

$$A^2m+B^3n=1-2mn \tag5$$

It is apparent that $m$ and $n$ do not have the same parity.

We have the following dependence:

$$(A^2+n_2)=(A+N_2)^2 \implies n_2 = N_2 (2 A + N_2) \implies n=2N(N+A) \tag6$$

It can be seen that $n$ must be even. As a result, $m$ must be odd. Equation $(6)$ may also be solved for $N$:

$$N=n-2A \tag7$$

We may now solve equation $(2)$ for $M$:

$$ \frac{\sqrt[3]{(A^2 B^3 + 2) (A + 2 N)}}{2 (A + 2 N)}-\frac{B}{2} \tag8$$

We may insert the result of equation $(7)$ into equation $(8)$, and the result is

$$M = \frac{\sqrt[3]{(n - A)(A^2B^3+2)}}{2 (n - A)}-\frac{B}{2} \tag9$$

As a result of a comparison of equations $(8)$ and $(9)$, we may write the following equations:

$$n=2(A+N) \tag{10a}$$

$$n=-2N \tag{10b}$$

Because of equation $(6)$, we may write the following equation

$$2(A+N)=2N(A+N) \implies N=1 \tag{11a}$$

$$N=-(A+1) \tag{11b}$$

The result of equation $(11$b) is effectively the same as $N=1$. We also know now that $n=2A+2$ (this result comes from both $(11$a) and $(11$b)). We may also write the first term of the largest powerful number: $(A+2)^2=A^2+4A+4$. By inserting into equation $(2)$, we get

$$M = \frac{1 - B^3}{A^2 + 2} \tag{12}$$

We may now solve equation $(12)$ for $A^2$

$$A^2 = -\frac{B^3 + 2 M – 1}{M} \tag{13}$$

In equation $(13)$, the nominator is even and the denominator is odd. In such a case, the result is even. However, $A$ is odd. This leads to a contradiction with the parity of $A$. The parity of $M$ is determined to be odd in conjunction with the equation $(3)$. When $B^3=B=1 \implies M=0$ and $A=0$.

As a consequence, three consecutive powerful numbers do not exist. There is a risk that I made a mistake somewhere, because I am blind to my own mistakes (like people tend to be).

[Beckon]: Beckon, Edward (2019), ”On Consecutive Triples of Powerful Numbers”, Rose-Hulman Undergraduate Mathematic Journal, Vol. 20: Iss. 2, Article 3.

$\endgroup$
7
  • 3
    $\begingroup$ Link to "similar kind of question" $\endgroup$ May 26 at 22:14
  • 6
    $\begingroup$ Finding the mistake(s) in your work is your job. $\endgroup$ May 27 at 1:54
  • 3
    $\begingroup$ @GerryMyerson this is literally the point of this website. $\endgroup$
    – 2132123
    May 27 at 9:06
  • 5
    $\begingroup$ @213, it's one thing when a student wants help with a Calculus exercise. It's a very different thing when someone thinks he/she presents a "proof" of a notorious open question, using nothing beyond high school math. If you're going to do math research, you have to learn to do it the way mathematicians do it, which is to be your own fiercest critic, searching out the weakest points in your argument, finding your mistakes before anyone else does. Shooting down feeble attempts at major questions is definitely not the point of this website. $\endgroup$ May 27 at 11:09
  • 3
    $\begingroup$ @GerryMyerson He says he is a "non-mathematician." The point of the website, as I see it, is help people with math. You can keep your judgments to yourself whether it's a feeble attempt or not. $\endgroup$
    – 2132123
    May 27 at 14:30

1 Answer 1

12
$\begingroup$

Equation (7) is incorrect; it should read $2N=\frac nN-2A$, not $2N=n-2A$. Dropping this factor of $N$ leads to the deduction that $N=1$ -- without this deduction, the rest of the proof falls apart.

(One reason this proof must fail is that it seems to predict that no two consecutive odd numbers, i.e. odd numbers which differ by 2, may be powerful. There are quite a few examples of such numbers; the smallest pairs are $(25,27)$ and $(70225, 70227)$, where $70225=5^253^2$ and $70227=3^517^2$.)

$\endgroup$
7
  • 1
    $\begingroup$ Indeed. Solving for $N$ in $(6)$ requires solving a quadratic, from which we get: $$N = \frac{-A \pm \sqrt{A^2 +2n}}{2}$$ You could resubstitute $(A^2 +2n)=(A+2N)^2$ in there, from which you get $N = -A/2$ (the other "root" becomes $N=N$). The calculations afterward don't work without changing this. $\endgroup$ May 27 at 2:19
  • 1
    $\begingroup$ Oh, even worse, having $2N = -A$ causes a ton of divide by zero errors. That's what happens when you sub 1 for 2, solve a thing, then resub 2 in for 1... tautologies and falsehoods. $\endgroup$ May 27 at 2:27
  • 1
    $\begingroup$ Excellent comments again. Thanks for them. Please note an even number cannot be an odd number ($2N=-A$). $\endgroup$
    – Tanaka
    May 27 at 4:51
  • $\begingroup$ So, I went "wait what" about $3^5 17^2$, since the pattern was supposed to be $a^2 b^3$. But of course it can be arranged as $3^3 (17\cdot3)^2$. And one of $a$ or $b$ could be one as well. I guess I like the other definition more, that "for a powerful number $m$, for every prime number $p$ dividing $m$, $p^2$ also divides $m$". Not sure how common knowledge the definitions and their equivalence are. $\endgroup$
    – ilkkachu
    May 27 at 9:41
  • 1
    $\begingroup$ @EricSnyder, yep. I did get it eventually, it might just be showing I'm an engineer and didn't actually study math. I admit $A^2B^3; A,B \in \mathbb{N}$ is a concise way of putting it, but concise isn't always the easiest to understand. (I think I confused it with $p^2q^3$ (with p,q primes) at first.) $\endgroup$
    – ilkkachu
    May 31 at 11:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.