An Exponential Diophantine Equation | Number Theory

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Опубликовано:

13 июн 2024

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Комментарии : 16

@wtspman 14 дней назад

My thought was to convert 42 to binary = 101010. There are ones in three of the binary digits => those give you the values of a, b, and c. If there were ones in anything other than 3 of the binary digits then there would be no solution for the sum of three powers of 2.

@wafflaaar1067 14 дней назад

woah.. this is a high IQ solution.

@giorgiocanal1659 11 дней назад

Simply awesome!

@WahranRai 14 дней назад

Similar to binary conversion, successive subtraction of the nearest power of 2 42 - 2^5 (32) = 10 ➔ 10 - 2^3 (8) = 2 ➔ 2 - 2 ^1 (2) = 0 ➔ 42 = 32 + 8 + 2 = 2^5 + 2^3 + 2^1 ➔ {a,b,c} = {5,3,1}

@Kounomura 14 дней назад

The task can be completed by representing 42 in the number system of two. 42= 32+8+2=2^5+2^3+2^1 {a,b,c}={5,3,1}

@littlegamer944 12 дней назад

I have a question please solve this - 2520 = (x+y+xy)^2+2xy+2y-3x Find possible value of y if x and y are pos integers

@wafflaaar1067 14 дней назад

How I tried doing it: as 42 is a multiple of 2, I thought to myself: 42 = 2 * something in fact, we can factor out a 2 from the sum of powers of 2. we go from 2^a + 2^b + 2^c = 42 2 ( 2^(a-1) + 2^(b-1) + 2^(c-1) ) = 42 divide both sides by 2 2^(a-1) + 2^(b-1) + 2^(c-1) = 21 we have 21 on RHS and it is odd... but we have powers of 2. only odd power of 2 is 1, therefore let 2*(c-1) = 1 , c - 1 = 0, c = 1 giving... 2^(a-1) + 2^(b-1) + 1 = 21 subtract 1 frm both sides -> 2^(a-1) + 2^(b-1) = 20 now, which powers of 2 give us 20? 16 and 4 (the others... well, they dont. 8 needs 12 to make 20 but 12 is not a power of 2, 2 needs 18 to make 20, but 18 not a power of 2) let 16 = 2^(a-1), 4 = 2^(b-1) therefore a = 5, b = 3, c = 1

@jaimeduncan6167 14 дней назад

I did it almost like you, once you have the 20 you can factor 20 as 2^2*5 divided by 2^2 and apply one more time the first step because again, 5 is odd. Then you get the response.I prefer it like that because it's more systemic, like, beautiful.

@prasadbmvs 11 дней назад

Nice explanation

@FisicTrapella 14 дней назад

2^a(1 + 2^(b-a) + 2^(c-a)) = 2·3·7 So, a=1 and 2^(b-1) + 2^(c-1) = 20 That is, you need 2 powers of 2 whose sum equals 20 and the only case is 16+4. So, b=5 and c=3.