The conjecture on digital anomalies with exponent two

From papers

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

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

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

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

Progress summary

Open

The conjecture remains open: the paper lists two examples but gives neither a proof that they are the only ones nor a counterexample.

The conjecture asserts that the two listed quadruples exhaust digital anomalies with exponent k=2k=2. It appears as Conjecture 4.6 in Samer Seraj’s paper, published in December 2025.

Known results

  • Assuming the abcabc conjecture, the paper proves finiteness for each fixed k3k \ge 3; this argument does not address k=2k=2.

December 2025 formulation

Seraj explicitly presents the k=2k=2 classification as a conjecture and says the abcabc approach does not appear feasible because the required positive exponent is unavailable. No retrieved source reports a proof, counterexample, verification, or retraction.

Current status (as of August 2026): the two listed examples are recorded, but the k=2k=2 classification remains unproved and no verified counterexample is reported.

Sources
Sources & referencesView supporting material

Primary source

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

Solutions 1

Counterexample

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

For each integer m1m\ge1, define positive integers am,bma_m,b_m by

bm+am2=(3+22)m.b_m+a_m\sqrt2=(3+2\sqrt2)^m.

Taking norms gives Pell's equation

bm22am2=1.b_m^2-2a_m^2=1.

Every ama_m is even: a1=2a_1=2, and

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

Now set

ym=am2,Bm=ambm,xm=am2bm22=Bm22.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:

xmym=bm22=am2+12=ym+xmBm2.\frac{x_m}{y_m} =\frac{b_m^2}{2} =a_m^2+\frac12 =y_m+\frac{x_m}{B_m^2}.

Moreover, since Bm6B_m\ge6,

Bmxm=Bm22<Bm2.B_m\le x_m=\frac{B_m^2}{2}<B_m^2.

Thus xmx_m has exactly two base-BmB_m digits, and

(xm,ym,Bm,2)(x_m,y_m,B_m,2)

is a digital anomaly for every m1m\ge1. The Pell solutions are strictly increasing, so these anomalies are pairwise distinct.

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

(18,4,6,2).(18,4,6,2).

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

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

Indeed,

20808144=144+12=144+208082042,20420808<2042=41616.\frac{20808}{144} =144+\frac12 =144+\frac{20808}{204^2}, \qquad 204\le20808<204^2=41616.

Consequently, the conjectured classification omits infinitely many solutions.

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Shivam Patel ·