Solutions to Diophantine Equation x (x + 1) = y (y + 5) (y + 10) (y + 15)
Theorem
- $n = x \paren {x + 1} = y \paren {y + 5} \paren {y + 10} \paren {y + 15}$
has exactly $2$ solutions in $\N \setminus \set 0$:
| \(\ds 1056\) | \(=\) | \(\ds 32 \times 33 = 1 \times 6 \times 11 \times 16\) | ||||||||||||
| \(\ds 43 \, 056\) | \(=\) | \(\ds 207 \times 208 = 8 \times 13 \times 18 \times 23\) |
Proof
First, we observe that:
| \(\ds x \paren {x + 1}\) | \(=\) | \(\ds \paren {x + \dfrac 1 2 }^2 - \paren {\dfrac 1 2}^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \paren {x + 1}\) | \(\lt\) | \(\ds \paren {x + \dfrac 1 2 }^2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt {x \paren {x + 1} }\) | \(\lt\) | \(\ds \paren {x + \dfrac 1 2 }\) | taking the square root of both sides |
We note here that the square root of the product $x \paren {x + 1}$ has a fractional part which is less than one half.
Therefore, the fractional part of the square root of the product of the four integers on the right hand side must also be less than one half.
Next, we observe that:
| \(\ds y \paren {y + 5} \paren {y + 10} \paren {y + 15}\) | \(=\) | \(\ds \paren {y^2 + 15 y} \paren {y^2 + 15 y + 50}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {y^2 + 15 y} \paren {y^2 + 15 y + 25} + 25 \paren {y^2 + 15 y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {y^2 + 15 y} \paren {y^2 + 15 y + 25} + 25 \paren {y^2 + 15 y} + 25^2 - 25^2\) | adding $0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {y^2 + 15 y} \paren {y^2 + 15 y + 25} + 25 \paren {y^2 + 15 y + 25} - 25^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {y^2 + 15 y + 25}^2 - 25^2\) |
Let $a = \paren {y^2 + 15 y + 25}$
We know that $\sqrt {a^2 - 25^2 } \lt a$
We can narrow the search for solutions tremendously as follows:
We need $\sqrt {a^2 - 25^2 } \lt a - \dfrac 1 2$ so that the fractional part of $\sqrt {a^2 - 25^2 }$ might be less than one half.
Therefore:
| \(\ds \sqrt {a^2 - 25^2 }\) | \(\lt\) | \(\ds a - \dfrac 1 2\) | the fractional part of the square root of the product must be less than one half. | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2 - 625\) | \(\lt\) | \(\ds a^2 - a + \dfrac 1 4\) | squaring both sides | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a\) | \(\lt\) | \(\ds \dfrac {2501} 4\) |
Therefore:
| \(\ds \paren {y^2 + 15 y + 25}\) | \(\lt\) | \(\ds \dfrac {2501} 4\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {y^2 + 15 y + 25} - \dfrac {2501} 4\) | \(\lt\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {y^2 + 15 y - \dfrac {2401} 4 }\) | \(\lt\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\lt\) | \(\ds \dfrac {-15 \pm \sqrt {15^2 - 4 \times \paren {-\dfrac {2401} 4} } } 2\) | Quadratic Formula | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\lt\) | \(\ds \dfrac {\sqrt {2626 } - 15 } 2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\le\) | \(\ds 18\) |
We now only need to review $18$ cases which are shown below.
$\blacksquare$
Historical Note
According to David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$, this result appeared in Acta Arithmetica on page $194$ of volume $7$.
However, this contributor has tried to identify the article in question, and it appears to have nothing to do with the theorem discussed here.
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1056$