As we have seen, all these will be squares if

s2 = x2 + y2 + 1 ± 2 √(x2y2 + x2 + yo).

We have also seen (Problem (1) above) that x+x+y becomes a square if only y=x+1. Put then y=x+1, when we have

that is,

g2 = 2x2 + 2x + 2 ± 2 √(x2 + 2x3 + 3x2 + 2x + 1) ;

z3 = 4 (x2 + x + 1).

It only remains to make x+x+1 a square. Equate this to (−x+1)3, and we have

Let q = 1, r = 2, and we have x, y = {, s= }; or, if we put = 2 in the values expressed in terms of t, the values are x = }, y = †, z = —

PROBLEM 17. To find two fourth powers A', B' such that their sum is equal to the sum of two other fourth powers'.

In other words, to solve the equation A‘+ B* = C‘+ D', or (what is the same thing) A' - D` = C‘— B‘.

It is proved, says Euler, that the sum of two fourth powers cannot be a fourth power, and it is confidently afirmed that the sum of three fourth powers cannot be a fourth power. But the equation A‘+ B' – C` = D' is not impossible.

1 Novi Commentarii Acad. Petropol., 1772, Vol. XVII. (1773), pp. 64 sqq.=Commentationes arithmeticae, 1. pp. 473-6; Mémoires de l'Acad. Imp. de St Pétersbourg, XI. (1830) pp. 49 sq. Comment. arithm., II. pp. 450-6.

=

One obvious case is obtained by putting k = ab, for then

whence y = a, x = 1, so that p = a, q = ab, r = ab, s = a, and the result is only the obvious case where p =s, q = r.

Following up this case, however, let us put kab (1 + 5).

therefore, multiplying numerator and denominator by 1-s and extracting the square root, we obtain

a √{(ba − 1)2 + ( 362 — 1) (1 − 1 ) 5 + 36a (¿a − 2) sa + ba (ba — 4) .~o — b31⁄23}

To make the expression under the radical a square, equate it to

and assume ƒ, g such that the terms in z, z vanish.

In order that the term in z may vanish, ƒ= }} (31⁄23 — 1), and, in order that the term in z may disappear,

36a (6a — 2) = 2 (b2 − 1 ) g + ƒ2 = 2 (b2 − 1 ) 8 + † (981 — 66a + 1),

Now b can be chosen arbitrarily; and, when we have chosen it and thence determined z, we can put

q=ab (b2 - 1+fz+gz3), s = a (b2 − 1 + fx + gx3),

where we may also divide out by a.

If x, y have a common factor, we may suppose this eliminated before P, 9, r, s are determined.

4

Let b= 2 (for b cannot be 1, since then g would be ∞).
f=, 8=-, *=000.

Ex. I.
Therefore

6600

As a does not enter into the calculation, we may write for it;

Ex. a. Let 63; therefore f= 13, 8 = 4, s = 188;

Another solution in smaller numbers.

In the second of the papers quoted Euler says that, while investigating quite different matters, he accidentally came across four much smaller numbers satisfying the conditions, namely,

A 542, B103, C=359, D= 514,

which are such that A+ B* = C' + D'.

He then develops two methods of analysis leading to this particular solution; but, while they illustrate the extraordinary ingenuity which he brought to bear on such problems, they are perhaps of less general interest than the above.

INDEX.

[The references are to pages.]

I. GREEK.

duvaμókußos, "square-cube" (=xo), sign ・ μelzwa ñ ¿» Xbyw, 132 N., 144 m.