[math-fun] numerical curiosity
bonjour tout le monde, Here is a numerical curiosity: 588^2 + 2353^2 = 5882353 the curious thing is that 1/17 = 0.05882353 (rounded to 7 digits). I never searched beyond that, are there other known results in this direction? Simon Plouffe
Here is a numerical curiosity: 588^2 + 2353^2 = 5882353 the curious thing is that 1/17 = 0.05882353 (rounded to 7 digits). ... are there other known results in this direction?
Yes. If A^2+1 divides B^A-A, then in base B, 1/(A^2+1) has that curious property. Examples: 3^2+1 divides 7^3-3, so in base 7: 1/(3^2+1) ~= .046_205, and 46^2+205^2=46205. 4^2+1 divides 6^4-4, so in base 6: 1/(4^2+1) ~= .0204_1225, and 204^2+1225^2=2041225. 5^2+1 divides 5^5-5, so in base 5: 1/(10^2+1) ~= .00440_04401, and 440^2+4401^2=44004401. 7^2+1 divides 43^7-7, so in base 43: 1/(7^2+1) ~= .(00)(36)(42)(06)(00)(36)(42)_ (06)(00)(36)(42)(06)(00)(37), and this one works, too. In each case, A says how many digits to take, and you round upward. I don't know of another base-ten instance. -- Don Reble djr@nk.ca
Simon Plouffe asked:
Here is a numerical curiosity: 588^2 + 2353^2 = 5882353 the curious thing is that 1/17 = 0.05882353 (rounded to 7 digits). ... are there other known results in this direction?
Let A, M, and D satisfy [1] M = A^2+1; B^D == A (mod M). and let [2] X = (B^D-A)/M; Y = (AB^D+1)/M; Z = (B^(2D)+1)/M. we can then show [3] Z = X^2+Y^2 = (B^D)X+Y. which is to say, Z = X^2+Y^2 is the numerical value of the concatenation of the base-B representations of X and Y, each left-zero-padded to D digits. Also, if Z-1 is converted to base B and left-zero-padded to 2D digits, the result is the block of 2D digits following the point in the base-B representation of 1/M. --------------------------------------------------------- Plouffe's observation X = 588, Y = 2353 corresponds to the solution B = 10, A = 4, M = 17, D = 4 of [1]. In a subsequent note, Reble gives additional examples corresponding to the following solutions to [1]: B = 7, A = 3, M = 10, D = 3 B = 6, A = 4, M = 17, D = 4 B = 5, A = 5, M = 26, D = 5 B = 43, A = 7, M = 50, D = 7 These solutions prove that Plouffe's observation is not a fluke; it can be generalized. However, Reble, apparently influenced by Plouffe's solution, made the unnecessary assumption A = D, which prevented him from finding additional solutions with B = 10. --------------------------------------------------------- Actually, there are an (apparent) infinitude of solutions with distinct M to [1] for B = 10. The smallest solutions in terms of M are: A = 4, M = 17, D = 4 A = 10, M = 101, D = 1 A = 16, M = 257, D = 64 A = 20, M = 401, D = 50 A = 24, M = 577, D = 432 A = 36, M = 1297, D = 972 A = 40, M = 1601, D = 150 A = 56, M = 3137, D = 784 A = 84, M = 7057, D = 1764 A = 90, M = 8101, D = 405 The first solution is Plouffe's observation, the second is a degenerate solution 0^2+0^2 = 00, the remainder are new (to me). Computer search generates further solutions ad nauseum. Also, a single solution for a given B and M seems to imply an infinite family of solutions with distinct D (including the degenerate solution above). For example, Plouffe's solution generalizes to B = 10, A = 4, M = 17, D = 16K+4 (K >= 0). The special case K = 1 yields 5882352941176470588^2 + 23529411764705882353^2 = 588235294117647058823529411764705882353. --------------------------------------------------------- The raw solution 0588^2 + 2353^2 = 05882353 remains valid when leading zeroes are removed, that is 588^2 + 2353^2 = 5882353. Let's call such a solution "desirable". Under this definition, the base 5 solution given by Reble: 00440^2 + 04401^2 = 0044004401 (base 5) is undesirable in that it becomes false when leading zeroes are stripped: 440^2 + 4401^2 = 44004401 (base 5). A solution is desirable iff the base-B representation of Y in [2] has D digits, that is Y >= B^(D-1). If we follow through the math, we find this happens only when A < B. For B = 10, the only solutions with A < B are those for which A = 4. Hence, the only desirable base-10 solutions* are in the family B = 10, A = 4, M = 17, D = 16K+4 (K >= 0). *We might call these "reciprocal type solutions" to x^2+y^2 = x concat y. There are other solutions which do not seem to be reciprocal type solutions, such as 10^2+1^2 = 101 12^2+33^2 = 1233 10^2+100^2 = 10100 88^2+33^2 = 8833 990^2+100^2 = 990100 etc.
participants (3)
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David Wilson -
Don Reble -
Simon Plouffe