3/28/2023 0 Comments Polyroots nspireThe sum is ends up a multiple of 111, but not necessarily of 11. We can construct two similar sums for BCA and CAB. Let a, b, and c represent the digits A, B, and C, respectively. Can we show that this is true for all sums AB + BA? I was unsuccessful.Īll the sums for this group are multiples of 11. My original goal was to find a number AB such that AB + BA = n where n contains both digits A and B. Let AB represent a two digit number, where A and B each represent a digit 0 through 9 (excluding 00). It is also an inspiration for me to blog. Please take a look at the Fun With Num3ers blog by Benjamin Vitale (Twitter: This is a fun blog working with number theory. Happy Holidays everyone and see you next time! Success! I believe x=1 and n=2 is the only solution to this equation with these conditions imposed. There are no solutions (in the natural number set) for x + x^2 + x^3 = n^2.Īgain, the only possible candidate is when x=1, but that leaves n = ∛3. And 1 * √3 = √3, n = √3, which is not a natural number. Once again the only way 1/x + 1 + x is an integer is that when x = 1. Can we find solutions with these conditions? (II) does not fit because 1 * √2 = √2, leaving n = √2, not fitting the requirement that n,x ∈ N.Īccording to this analysis, there are no natural number solutions to x + x^2 = n^2. We know that 2 is not a perfect square (√2 ≈ 1.41421), so condition 2 fails. The only natural number that allows condition 1 to be true is when x = 1. The quantity 1/x + 1 has to be an integer,Ģ. Taking the (principal) square root of both sides yields:ġ. Our first instinct is most likely to go grab the nearest calculator or computer. Some mathematicians include 0.Īre there any integer solutions to x + x^2 = n^2 with n,x ∈ N? Natural numbers are commonly referred to the whole numbers 1, 2, 3, etc. And yes, I am 99.44% confident that the human race will still be on Earth be here come. Hi everyone! Hopefully you are fine today. The solution to a^2 + b^2 = 2*a*b is when a = b This makes sense because when a=b, a^2 + a^2 = 2 * a * a = 2 * a^2. One solution to a + b^2 = a * b is a=4 and b=2. If a^2 - 4a is a prefect square, then a^2 - 4a = n^2. If a = 16, then a^2 - 4a = 192, not a perfect square. If a = 14, then a^2 - 4a = 140, not a perfect square. If a = 12, then a^2 - 4a = 96, not a perfect square. If a = 10, then a^2 - 4a = 60, not a perfect square. If a = 8, then a^2 - 4a = 32, not a perfect square. If a = 6, then a^2 - 4a = 12, not a perfect square. Not good for searching for solutions such that a,b ∈ N.Ī = (b ± √(b^2 - 4b^2))/2 = (b ± i b √3)/2 = b * (1 ± i √3)/2 which is not a natural number. This lead me to believe that b is a multiple of a. Let's start with subtracting a*b from both sides: What if we allow the possibility that a ≠ b? The only solution to a + b = a * b is a = b = 2. The only way that 1/(b - 1) ∈ N is when b = 2. This blog entry is all about trying to find solutions to equations - in an analytic manner. Note that the set of natural numbers may or may not include 0. That is all variables are natural numbers (the counting numbers). Note I am looking for integer solutions, ignoring all solutions that were not natural numbers.įor the following let a, b, and n ∈ N. Example: int(π) = 3.īelow is the instructions on how I found possible solutions on the Casio Prizm. The integer function returns the integer part of a number. Then solving for A would yield A = √(X³ - B²). I will post screen shots from a Casio Prizm to demonstrate the search for solutions.įor each case, I will present base cases (A = 0, B = 0, and A = B = 0). Type each command as you would entering commands in Program mode. On the current generation (fx-9860g II, Prizm, this includes the older models), within the RUN/MAT mode set the calculator in Linear Input mode. One great thing about Casio graphing calculators is that you can make mini-programs. The set can include or not include zero, depending in the definition used. (Where N is the set of natural numbers (counting numbers). Here's to a Happy and Safe New Years Celebration and to a kick ass 2013! To my readers, I get such pleasure from writing on this blog, and fully appreciate each and every one of you.
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