A digital anomaly is a quadruple (x,y,B,k) satisfying the paper's definition. In the case k=2, the conjecture concerns the complete list of such quadruples. Digital-anomaly conjecture for k=2. The only digital anomalies (x,y,B,k) with k=2 are (x,y,B,k)=(18,4,6,2)quadandquad(1323,36,42,2). The conjecture is proposed because the abc conjecture does not appear to yield the required positive exponent for the k=2 case. Its resolution status is not specified in the source.

The conjecture on digital anomalies with exponent two

A digital anomaly is a quadruple $(x,y,B,k)$ satisfying the paper's definition. In the case $k=2$ , the conjecture concerns the complete list of such quadruples.

Digital-anomaly conjecture for $k=2$ . The only digital anomalies $(x,y,B,k)$ with $k=2$ are

(x,y,B,k)=(18,4,6,2)\quad\text{and}\quad(1323,36,42,2).

The conjecture is proposed because the $abc$ conjecture does not appear to yield the required positive exponent for the $k=2$ case. Its resolution status is not specified in the source.

References

Primary source

Samer Seraj, “Diophantine Analysis of a Digital Anomaly”, arXiv:2512.06056 (2025).

diophantine-equations number-theory

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-proposed infinite family would refute the conjecture, but the claim has not been independently verified.

Samer Seraj proposed in 2025 that the two listed quadruples are the only digital anomalies with exponent $2$ . The paper labels this a conjecture, not a theorem, and notes that the $abc$ approach does not handle this case.

Known results

Assuming the $abc$ conjecture, Seraj proves finiteness for each fixed $k\ge3$ ; this does not settle $k=2$ .

Posted attempt

A reader claims an infinite Pell-equation family of counterexamples, including $(20808,144,204,2)$ , so the proposed classification is false. The attempt claims a complete refutation, but it has not been independently verified.

Current status (as of August 2026): a posted, unverified construction claims to refute the $k=2$ classification; absent verification, the conjecture is not settled.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

The proposed two-element list is false. In fact, there are infinitely many digital anomalies with exponent $k=2$ .

For each integer $m\ge1$ , define positive integers $a_m,b_m$ by

b_m+a_m\sqrt2=(3+2\sqrt2)^m.

Taking norms gives Pell's equation

b_m^2-2a_m^2=1.

Every $a_m$ is even: $a_1=2$ , and

a_{m+1}=2b_m+3a_m.

Now set

y_m=a_m^2,\qquad B_m=a_mb_m,\qquad x_m=\frac{a_m^2b_m^2}{2} =\frac{B_m^2}{2}.

These are positive integers, and the defining digital-anomaly equation holds:

\frac{x_m}{y_m} =\frac{b_m^2}{2} =a_m^2+\frac12 =y_m+\frac{x_m}{B_m^2}.

Moreover, since $B_m\ge6$ ,

B_m\le x_m=\frac{B_m^2}{2}<B_m^2.

Thus $x_m$ has exactly two base- $B_m$ digits, and

(x_m,y_m,B_m,2)

is a digital anomaly for every $m\ge1$ . The Pell solutions are strictly increasing, so these anomalies are pairwise distinct.

For $m=1$ , this construction recovers the known example

(18,4,6,2).

For $m=2$ , it produces the unlisted counterexample

\boxed{(20808,144,204,2).}

Indeed,

\frac{20808}{144} =144+\frac12 =144+\frac{20808}{204^2}, \qquad 204\le20808<204^2=41616.

Consequently, the conjectured classification omits infinitely many solutions.