CHAP. IV] PROBLEMS ON THE SIDES OF RIGHT TRIANGLES. 187 



Saint-Martin asked how many ways 1803601800 is the difference of the 

 [larger]] sides of a right triangle whose least side differs from the other 

 sides by squares. Fermat 145 replied that there are exactly 243 such triangles. 



Fermat 146 asked for two right triangles such that the product of the 

 hypotenuse and least leg of one shall have a given ratio to the corresponding 

 product for the other triangle. 



Under 2x 4 y* = D in Ch. XXII are discussed right triangles whose 

 hypotenuse is a square and either the sum of the legs is a square or the 

 least side differs by a square from each of the remaining sides. 



Fermat 147 gave (156, 1517, 1525) in reply to Frenicle's question to find 

 a right triangle in which the square of the difference of the legs exceeds the 

 double of the square of the least leg by a square. A. Aubry 148 obtained an 

 infinity of solutions by descent. 



Frenicle 89 noted (pp. 71-8) that if the hypotenuse and perimeter of a 

 right triangle are squares, the perimeter has at least 13 digits. 



J. Ozanam 149 gave a rule to find a right triangle whose hypotenuse exceeds 

 the larger leg by unity [Pythagoras 1 ]. From the lengths of its legs form a 

 new triangle; its hypotenuse is a square. He found right triangles whose 

 base and hypotenuse are triangular numbers and altitude is a cube. 



Wm. Wright 150 found a right triangle the sum of whose perimeter and 

 square of any side is a square. Let the sides be ax, bx, ex, where a 2 + 6 2 = c 2 . 

 Then x 2 + px, x 2 + qx, x 2 + rx are made squares in the usual way (Ch. 

 XVIII), where p = s/a 2 , q = s/b 2 , r = s/c 2 , s = a + b + c. He and others 151 

 gave a similar treatment to find a right triangle such that the square of 

 any side exceeds that side by a square. 



Several 152 found a right triangle whose perimeter is a square, also the 

 sum of the square of any side and the remaining two sides, also the sum of 

 any side and the square of the sum of the remaining two sides. These seven 

 conditions are satisfied if the sum of the sides is 1/4. Take/(p 2 T q 2 ), 2fpq 

 as the sides. Equating the sum to 1/4, we get /. 



R. Tucker and S. Bills 153 found a right triangle with perimeter a square 

 and diameter of the inscribed circle a cube [or vice versa]. Let the sides 

 be (p 2 q 2 }x, 2pqx. Then 2p(p + q)x = D = r 2 , and the diameter 

 2q(p q)x is to be a cube, say r 3 /s 3 . From the two values of x we get r 

 in terms of q, s. 



A. B. Evans 1530 found a right triangle with integral values for the sides 

 a, b, c, diameter d of the inscribed circle and side s of the inscribed square 

 having one angle coincident with the right angle of the triangle and having 



145 Oeuvres, II, 250, letter to Mersenne, Jan. 27, 1643. 



146 Oeuvres, II, 252, letter to Mersenne, Feb. 16, 1643. 



147 Oeuvres, II, 265, letter to Carcavi, 1644. 



148 L'intermediaire des math., 20, 1913, 141-4. 



149 Recreations Math., 1, 1723, 1735, 52-5. 



160 The Gentleman's Math. Companion, London, 5, No. 24, 1821, 59-60. 



161 Ibid., 5, No. 27, 1824, 312-6. 



162 Ibid., 5, No. 25, 1822, 157-9. 



153 Math. Quest. Educ. Times, 19, 1873, 82. 

 i B3 Ibid., 21, 1874, 103-4. 



