---

Statement

Example (p. 116, unnumbered). The equation $7X^2-3Y^2=1$ has smallest solution $2\sqrt7+3\sqrt3$, and

$$(2\sqrt7+3\sqrt3)^7=2{,}637{,}362\sqrt7+4{,}028{,}637\sqrt3 ;$$

this solution and all its odd powers have property $Q$, that is, $7$ divides the coefficient of $\sqrt7$ and $3$ divides that of $\sqrt3$. The seventh power gives the consecutive powerful numbers, neither a square,

$$48{,}689{,}748{,}233{,}308=7\cdot2{,}637{,}362^2 =2^2\cdot7^3\cdot13^2\cdot43^2\cdot337^2$$

and

$$48{,}689{,}748{,}233{,}307=3\cdot4{,}028{,}637^2 =3^3\cdot139^2\cdot9661^2 .$$

The exponent $7$ is the one Theorem 3.2 (3) prescribes, $7$ being the only odd prime dividing $mn=21$ but not $xy=6$; by Theorem 3.5 the odd powers of this solution are all the solutions with property $Q$, so the equation gives infinitely many such pairs. Writing $X=7x$ and $Y=3y$, they are the solutions of $7^3x^2-3^3y^2=1$.

The pair is the least one this equation gives, not the least consecutive powerful pair with neither member a square: Golomb's $12167=23^3$ and $12168=2^3\cdot3^2\cdot13^2$ is smaller (an observation of this page).

Source. D. T. Walker, Consecutive integer pairs of powerful numbers and related Diophantine equations, Fibonacci Quart. 14 (1976), no. 2, 111-116: the example on p. 116. The edition is identified on the source card.

Read depth. Claims checked: the example was read on the printed page and its arithmetic (the seventh power, both factorizations and the difference $1$) was checked here.

Bears on

• Problem 365: it gives infinitely many pairs of consecutive powerful numbers with neither member a square, so the first question, read as whether one member must be a square, has answer no. It gives no count of pairs up to $x$. The claim page Walker 1976 records this result.