DIOPHANTINE, 
Pros. XXIX. 
46. To find three fquare numbers in harmonical Bere 
tion, 4. ¢- to fip 2, as when as 
Be 
—a. Afflume two known igus for ie and a’, 
foppoe 49 and 25, then as x7: 25 : — 49 3 24, or 
x°—1225, or x == 1225. 
Pros. XXX. vide Pros. XXVIII. 
Firft find three fquare numbers in arithmetical pro- 
greflion, each of which wil! be the fum of two of the num- 
bers fought, and have the fame common difference as the 
numbers fought 
Thus let x*, x* + 20 y,and x* + 40 y, denote three {quare 
numbers in caine progreffion. Then diff. = (25.) 
zoy, the factors 2 y and io, the J diff. of them is y—5 = x. 
alias cae} =y?—loy + 25 x7 + 20 
Ioy +2 a iquare; but # arn + 30y + 25 is 
alfo to be made a {quare. 
Affume y+al? = 9? + 2ay +73 307 + 25= 
a@—25 
a’, i in 
as pe 
2ay + 
ais evidently 
ae between 5and15. Leta = 13, theny = 36, 20y= 
n diff. I. So tha ‘ 961 = ee 
= {quare of 49, are 
the three {quares ee and from the fum 961, and diff. 
720, we find 120%, 8403, and 15604, the eee nume 
bers required, Or, by denoting the three required {quares 
by x, x7 + 409s and x*+ 80, and proceeding as above, we 
a’? — 100 
have y = ae 
4 between 10 and 30; ifa = 26, then 
y = 72, and 4oy= 2880, and x = 62, and the three {quares, 
as well as the three numbers eee = 4 times thofe abov ove 
the = pofitive integral anfwe 
ore general folution may be had by ome 
vena {quares thus; x°, x° + gay, and x* + 8ay, then 
yx + Oay + @ is to be made a {quare, aflume = J +mal” 
a—a?* 
= y? +2 may + ma’, and we hall have y = = 
2 mm — 
PE Nad 5 , where m mutt evidently be between 
6—2m O—~2m 
I 
rand3. Ifm= = and a= 5, then y==-36, x = 31,&c. 
as above. 
Pros. XXXI. 
48. To find in whole — the three fides of a triangle = 
having one angle = 120. te VI. Geometry, fs: 84. 
ma. When ye i] = 120° 2 ide Cra’+ab+ 
&°. When 2 B= 60°, then c? = @? b+ 0% 
Demonfiration. 1, In a 'B ercducel take BD= BC 
=CD.fince 2 C B D= 60° Draw the dag! eae CE. 
then BE= 446= iBC; but 47. E?+ EC 
er oish a cies or EC? = 3b Now AE = 
4 
a+7), and AE? =a@+4ab64+i8.47.E.1. AC 
abd + B. 
=e’ ma 2 
= =a + 
Qe reg let AD =a, &c. as before; then AE = 
me ae ne ae or AC= eae 
ag + 2. E 
Cn. 3.—If 2; b by be the fides of a triangle, and 4 and 
b contain an angle of 120°; then a + 4, 6 and ¢ are the 
fides of a triangle, and a + 4 and 4 contain an angle of 
60° = ADC; anda+d,a “ c are a fides of a ae 
gle, agente d a contain an angie o Cz 
—The oO is true of ae eadmlcak: of a, b, "and 
¢, as ae bd, cd, & 
Solutio ume any number 4 for one of the 
taining fides, and put x = the ia then ae {quare of the 
third fide is 07> -+ Sx + x* perlemm 
Affume it = a— Pee — 200 $ #5 ore — 20x = 
te. lo 
&+bx,or2zax+bn= a — B, and x =< Pavey) Theres 
fore the three fides are 4, one. and a— 2. or 
b 2a+o 
2ab+Bh &—db 
zat aye Or (48. 4.) 2 a4 
ui 6, at — 8, and a + a6 + be are the three fides, all in- 
ers. 
we Pinca dhe ee 2ab6 + #,and2a$h 
P=s— a’, obi @+t+ab+P= -s — ab; whence this 
any numbers, a b. whofe fum is s, then 
‘the Saag fides are a? — 3°, s* — a, and the oppo- 
fite fide — 
ee = 2,0 = 1, and £ = 3; > 5. and 
4, are the deer or by aking of fum os we leait fides 
+5= have 5, fay Ss rthe fides of 
two ead triangles, having. ie ang ale or chie the fide 
Morec ver, the fum of the fquares +, the produ of the 
containing fides, is a {quare. (per 48.) 
Pros. XXXII. 
oO 
49. To find in whole numbers, three right-angled trian- 
gles, having all the fame area: the hypothenufe is out of the 
vefti 
2 
1, Find (per 48.) se “ fides & : and ¢, of a triangle, 
ai the ila oppolite ¢ = 120°, then of a and c, bard 
a ae fie ight triangles (per 21.) 
ay ag will all tae the fame a 
Demonflration, 2. The Canes formed of a and ¢ wiil 
have c? — a’, and 2 ac for its fides, and . 
itsarea, But (per 48.) c? = a? 4+ ab 
+h=—=b.atbhandac.c—-@oach.a+b = area 
of rft triangle. 
3: The fides of the 2d triangle, from 4 and c, arec? — J? 
ag and its area dc. 28 — o's but —& av+tab 
at+b..bc.0—B = bea. a+, the fame as 
ied area of the or jaa 
4. From a + 6 andc the fides ‘area + 0? —P =@ + 
2ab+ 8 ~a@+ab4P=abandrzac.a+dé. 
Ste its area is abc.a +, the fame as the other 
E 
two. Q. 
Fae —The numbers in 48. were 3, 5, 7. Of 3 and 
7 the triangle is go and 42, and area 840. of 5 and 7 the 
triangle is 24 and 70, and area 840. Of 8 and 7 the trians 
gle is 15 and 112, and area 840. 
Pros. XXXITI. 
50. To find three numbers a, 4, and ¢, fuch that s 
being th their fum a + 5, 6? + s, and ¢? + s, may be all fquare 
Q 
z. The 
