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In this expression x is an odd number prime to y; y may be odd or even; and neither is divisible by 5, as stated above. Let (x — y)/SH = d\ which for values satisfying (2) will be an integer. Evidently x and y must satisfy the congruence x^y (mod 5). Hence, for values of x up to any required magnitude, there are corresponding values of y; and the resultant values of d can be found and tabulated. If (2) is satisfied, then d=a+(3+y; and from the elementary properties of the sum of three biquadrates it is seen that many values of d may be rejected at once: for example, all those having for the final digit 0, 4, 5, or 9; and all those incapable of representing the sum of three square numbers. The remaining values of d will have for the final digit 1, 2, 3, 6, 7, or 8. Regarding these as separate cases, it is possible to devise a numerical test for each case. As an example, consider the values of d ending with • • • 3. These must have the form 80&+3 (sum of three odd biquadrates prime to 5) or the form 80&+33 (sum of one odd and two even biquadrates prime to 5). Choosing the latter form, seventy instances are found in a tabulation of values of d based on all admissible values of x <700. The first ten in order of magnitude are: 164833, 195313, 198593, 3029873, 4106193, 4590753, 5086913, 5693793, 5948193, 7424753. Testing these by a method (based upon the character of k), which need not be

A note on the diophantine problem of finding four biquadrates whose sum is a biquadrate