# Sums of Consecutive Sequences of Squares that equal Squares

## Theorem

The $24$th square pyramidal number is the only one which is square:

$1^2 + 2^2 + 3^2 + \cdots + 24^2 = 70^2$

while there are several Sum of Sequence of Squares which are square, for example:

$18^2 + 19^2 + \cdots + 28^2 = 77^2$

and:

$25^2 + 26^2 + \cdots + 624^2 = 9010^2$

## Proof

We have:

 $\ds 1^2 + 2^2 + 3^2 + \cdots + 24^2$ $=$ $\ds \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6$ Sum of Sequence of Squares $\ds$ $=$ $\ds \dfrac {24 \times 25 \times 49} 6$ $\ds$ $=$ $\ds \dfrac {2^3 \times 3 \times 5^2 \times 7^2} {2 \times 3}$ $\ds$ $=$ $\ds 2^2 \times 5^2 \times 7^2$ $\ds$ $=$ $\ds \paren {2 \times 5 \times 7}^2$ $\ds$ $=$ $\ds 70^2$

and:

 $\ds 18^2 + 19^2 + \cdots + 28^2$ $=$ $\ds \dfrac {28 \times \paren {28 + 1} \times \paren {2 \times 28 + 1} } 6 - \dfrac {17 \times \paren {17 + 1} \times \paren {2 \times 17 + 1} } 6$ Sum of Sequence of Squares $\ds$ $=$ $\ds \dfrac {28 \times 29 \times 57 - 17 \times 18 \times 35} 6$ $\ds$ $=$ $\ds \dfrac {\paren {2^2 \times 7} \times 29 \times \paren {3 \times 19} - 17 \times \paren {2 \times 3^2} \times \paren {5 \times 7} } {2 \times 3}$ $\ds$ $=$ $\ds 2 \times 7 \times 29 \times 19 - 17 \times 3 \times 5 \times 7$ $\ds$ $=$ $\ds 7 \times \paren {1102 - 255}$ $\ds$ $=$ $\ds 7 \times 847$ $\ds$ $=$ $\ds 7 \times 7 \times 11^2$ $\ds$ $=$ $\ds 77^2$

and:

 $\ds 25^2 + 26^2 + \cdots + 624^2$ $=$ $\ds \dfrac {624 \times \paren {624 + 1} \times \paren {2 \times 624 + 1} } 6 - \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6$ Sum of Sequence of Squares $\ds$ $=$ $\ds \dfrac {624 \times 625 \times 1249 - 24 \times 25 \times 49} 6$ $\ds$ $=$ $\ds \dfrac {\paren {2^4 \times 3 \times 13} \times 5^4 \times 1249 - \paren {2^3 \times 3} \times 5^2 \times 7^2} {2 \times 3}$ $\ds$ $=$ $\ds 2^3 \times 5^4 \times 13 \times 1249 - 2^2 \times 5^2 \times 7^2$ $\ds$ $=$ $\ds 2^2 \times 5^2 \times \paren {2 \times 5^2 \times 13 \times 1249 - 7^2}$ $\ds$ $=$ $\ds 2^2 \times 5^2 \times \paren {811 \, 850 - 49}$ $\ds$ $=$ $\ds 2^2 \times 5^2 \times \paren {811 \, 801}$ $\ds$ $=$ $\ds 2^2 \times 5^2 \times \paren {17^2 \times 53^2}$ $\ds$ $=$ $\ds 9010^2$

$\blacksquare$