gpt pro 5.4 attempt on Math-5606

Erdős Problem #365 --- ROUND 2

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction.

Round-1 correctly destroyed the first subquestion by one explicit counterexample, but it overclaimed a full resolution. The original problem has a second, still-open counting question: \[ \#\{n\le x : n\mbox{ and }n+1\mbox{ are powerful}\} \stackrel{?}{\ll} (\log x)^{O(1)}. \] So the most useful round-2 task is to correct the scope, and at the same time strengthen the first-part disproof from one counterexample to an infinite family.

2) Round-1 FOUNDATION USED

I use the following round-1 facts without reproving them.

• Round-1 isolated the precise first subquestion: if $n$ and $n+1$ are powerful, must one of them be a square?
• Round-1 gave the explicit counterexample \[ 12167=23^3,\qquad 12168=2^3\cdot 3^2\cdot 13^2, \] so the answer to that first subquestion is already no.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a self-contained Walker-type infinite family of counterexamples coming from the biquadratic unit \[ \beta:=2\sqrt{7}+3\sqrt{3}. \] Odd powers of $\beta^7$ produce infinitely many integer solutions to \[ 7^3x^2=3^3y^2+1, \] which in turn give infinitely many consecutive powerful pairs in which neither term is a square.

4) ATTACK PLAN (ROUND-2)

The exact round-1 gap is a scope gap, not a proof gap: the overall problem was marked "full solution" even though the counting question was untouched.

I will therefore do two things:

• prove an infinite family of counterexamples to the first subquestion;
• explain why the overall problem remains unresolved because the counting question survives unchanged.

5) WORK (ROUND-2)

Let \[ \beta:=2\sqrt{7}+3\sqrt{3},\qquad \beta^{\ast}:=2\sqrt{7}-3\sqrt{3}. \] Then \[ \beta\beta^{\ast}=28-27=1. \] A direct expansion gives \[ \beta^7=2637362\sqrt{7}+4028637\sqrt{3}. \] Set \[ \gamma:=\beta^7=a_1\sqrt{7}+b_1\sqrt{3}, \] so \[ a_1=2637362,\qquad b_1=4028637, \] and in particular \[ 7\mid a_1,\qquad 3\mid b_1. \]

For every odd integer $r\ge 1$, write \[ \gamma^r=a_r\sqrt{7}+b_r\sqrt{3} \] with integers $a_r,b_r>0$. (Odd powers have this form because multiplying an odd-form element $u\sqrt7+v\sqrt3$ by an even-form element $A+B\sqrt{21}$ again gives an odd-form element.)

I claim that for every odd $r$, \[ 7\mid a_r,\qquad 3\mid b_r. \] This is proved by induction on $r$ in steps of size $2$. The base case $r=1$ is already known. Now suppose \[ \gamma^r=A\sqrt7+B\sqrt3 \] with $7\mid A$ and $3\mid B$. Write \[ \gamma^2=u+v\sqrt{21} \] with integers $u,v$. Since \[ \gamma^2=(a_1\sqrt7+b_1\sqrt3)^2=(7a_1^2+3b_1^2)+2a_1b_1\sqrt{21}, \] we have \[ v=2a_1b_1, \] so $21\mid v$ because $7\mid a_1$ and $3\mid b_1$. Now \[ \gamma^{r+2}=\gamma^r\gamma^2=(A\sqrt7+B\sqrt3)(u+v\sqrt{21}) \] so \[ \gamma^{r+2}=(Au+3Bv)\sqrt7+(Bu+7Av)\sqrt3. \] Because $7\mid A$ and $7\mid v$, the coefficient $Au+3Bv$ is divisible by $7$. Because $3\mid B$ and $3\mid v$, the coefficient $Bu+7Av$ is divisible by $3$. Thus the divisibility property propagates from $r$ to $r+2$, proving the claim.

Next, conjugating in the $\sqrt3$-coordinate gives \[ \gamma^{\ast}:=a_1\sqrt7-b_1\sqrt3, \] and therefore for each odd $r$, \[ \gamma^r\,\gamma^{\ast r}=1. \] Since \[ \gamma^r=a_r\sqrt7+b_r\sqrt3, \qquad \gamma^{\ast r}=a_r\sqrt7-b_r\sqrt3, \] we obtain \[ 7a_r^2-3b_r^2=1. \] Because $7\mid a_r$ and $3\mid b_r$, the integers \[ x_r:=\frac{a_r}{7},\qquad y_r:=\frac{b_r}{3} \] are well-defined, and the identity becomes \[ 343x_r^2-27y_r^2=1. \] Hence the numbers \[ N_r:=27y_r^2, \qquad N_r+1=343x_r^2 \] are consecutive. Both are powerful:

• $27y_r^2=3^3y_r^2$ is powerful because all exponents coming from $y_r^2$ are even and the exponent of $3$ is at least $3$;
• $343x_r^2=7^3x_r^2$ is powerful for the same reason.

Neither is a square:

• the $3$-adic exponent of $27y_r^2$ is $3+2v_3(y_r)$, which is odd;
• the $7$-adic exponent of $343x_r^2$ is $3+2v_7(x_r)$, which is odd.

So every odd $r\ge 1$ gives a pair of consecutive powerful non-squares. Since there are infinitely many odd $r$, there are infinitely many such counterexamples.

The first member of this family comes from $r=1$: \[ x_1=376766,\qquad y_1=1342879, \] so \[ 27y_1^2=48689748233307, \qquad 343x_1^2=48689748233308. \] Their factorizations are \[ 48689748233307=3^3\cdot 139^2\cdot 9661^2, \] \[ 48689748233308=2^2\cdot 7^3\cdot 13^2\cdot 43^2\cdot 337^2. \] This is an explicit consecutive powerful pair in which neither term is a square.

Therefore the first subquestion is not merely false; it is false in an infinite family. However, the second counting question remains. So the original Erdős problem is still unresolved overall.

6) ADVERSARIAL VERIFICATION

• Hidden parity issue: only odd powers of $\gamma$ have the form $a\sqrt7+b\sqrt3$. I used only odd $r$ throughout.
• Integrality of $x_r,y_r$: proved from the inductive divisibility $7\mid a_r$ and $3\mid b_r$.
• Could one of the two numbers still be a square if $x_r$ or $y_r$ contains extra factors of $7$ or $3$? No. The exponents of $7$ and $3$ are respectively $3+2v_7(x_r)$ and $3+2v_3(y_r)$, always odd.
• Scope check: the original problem asks a second counting question, so even an infinite family of counterexamples to the first subquestion does not finish the whole problem.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The first subquestion is completely settled negatively, and round-2 strengthens round-1 from a single counterexample to an explicit infinite family. But the overall Erdős problem remains open because the counting question is still unresolved.

8) COMPLETION ESTIMATE

COMPLETION: 60%

9) REFERENCES

• Bloom365 T. F. Bloom, Erdős Problem #365, https://www.erdosproblems.com/365, accessed 2026-03-07.

---