Elements of Algebra

438 ELEMENTS PART II. found among those numbers; the property of $m^4 - n^4 = (m^2 + n^2) \times (m^2 - n^2)$, will enable us to construct for our purpose the following Table: A Table of Numbers contained in the Formula $m^4 - n^4$. | $m^2$ | $n^2$ | $m^2-n^2$ | $m^2+n^2$ | $m^4-n^4$ | |---|---|---|---|---| | 4 | 1 | 3 | 5 | $3 \times 5$ | | 9 | 1 | 8 | 10 | $16 \times 5$ | | 9 | 4 | 5 | 13 | $5 \times 13$ | | 16 | 1 | 15 | 17 | $3 \times 5 \times 17$ | | 16 | 9 | 7 | 25 | $25 \times 7$ | | 25 | 1 | 24 | 26 | $16 \times 3 \times 13$ | | 25 | 9 | 16 | 34 | $16 \times 2 \times 17$ | | 49 | 1 | 48 | 50 | $25 \times 16 \times 2 \times 3$ | | 49 | 16 | 33 | 65 | $3 \times 5 \times 11 \times 13$ | | 64 | 1 | 63 | 65 | $9 \times 5 \times 7 \times 13$ | | 81 | 49 | 32 | 130 | $64 \times 5 \times 13$ | | 121 | 4 | 117 | 125 | $25 \times 9 \times 5 \times 13$ | | 121 | 9 | 112 | 130 | $16 \times 2 \times 5 \times 7 \times 13$ | | 121 | 49 | 72 | 170 | $144 \times 5 \times 17$ | | 144 | 25 | 119 | 169 | $169 \times 7 \times 17$ | | 169 | 1 | 168 | 170 | $16 \times 3 \times 5 \times 7 \times 17$ | | 169 | 81 | 88 | 250 | $25 \times 16 \times 5 \times 11$ | | 225 | 64 | 161 | 289 | $289 \times 7 \times 23$ | We may already deduce some answers from this. For, if $f^2 = 9$, and $k^2 = 4$, we shall have $f^4 - k^4 = 13 \times 5$; farther, let $g^2 = 81$, and $h^2 = 49$, we shall then have $g^4 - h^4 = 64 \times 5 \times 13$; therefore $t^2 = 64 \times 25 \times 169$, and $t = 520$. Now, since $t^2 = 270400$, $f=3$, $g=9$, $k=2$, $h=7$, we shall have $a=21$, and $b=18$; so that $p=117$, $q=765$, and $r=756$; from which results $2x = t^2 + p^2 + q^2 = 869314$; consequently, $x=434657$; then $y = x - p^2 = 420968$, and lastly, $z = x - q^2 = -150568$. This last number may also be taken positively; the difference then becomes the sum, and, reciprocally, the sum becomes the difference. Since therefore the three numbers sought are: