2-digit num3ers expressible as 10*a + b – (a^2 + b^2)

To find a 2-digit number ab such that
ab   –   (a^2 + b^2)   =   N,    N   =   10,   11,   12,   13   …   ,   99

 
10   =   10*2   +   3   –   (2^2   +   3^2)
10   =   10*8   +   3   –   (8^2   +   3^2)

11   cannot be expressed

12   =   10*4   +   4   –   (4^2   +   4^2)
12   =   10*6   +   4   –   (6^2   +   4^2)

13   =   10*5   +   4   –   (5^2   +   4^2)

14   =   10*2   +   2   –   (2^2   +   2^2)
14   =   10*8   +   2   –   (8^2   +   2^2)

15   =   10*3   +   3   –   (3^2   +   3^2)
15   =   10*7   +   3   –   (7^2   +   3^2)

16   =   10*2   +   0   –   (2^2   +   0^2)
16   =   10*2   +   1   –   (2^2   +   1^2)
16   =   10*8   +   0   –   (8^2   +   0^2)
16   =   10*8   +   1   –   (8^2   +   1^2)

17   cannot be expressed

18   =   10*4   +   3   –   (4^2   +   3^2)
18   =   10*6   +   3   –   (6^2   +   3^2)

19   =   10*3   +   2   –   (3^2   +   2^2)
19   =   10*5   +   3   –   (5^2   +   3^2)
19   =   10*7   +   2   –   (7^2   +   2^2)

20   cannot be expressed

21   =   10*3   +   0   –   (3^2   +   0^2)
21   =   10*3   +   1   –   (3^2   +   1^2)
21   =   10*7   +   0   –   (7^2   +   0^2)
21   =   10*7   +   1   –   (7^2   +   1^2)

22   =   10*4   +   2   –   (4^2   +   2^2)
22   =   10*6   +   2   –   (6^2   +   2^2)

23   =   10*5   +   2   –   (5^2   +   2^2)

24   =   10*4   +   0   –   (4^2   +   0^2)
24   =   10*4   +   1   –   (4^2   +   1^2)
24   =   10*6   +   0   –   (6^2   +   0^2)
24   =   10*6   +   1   –   (6^2   +   1^2)

25   =   10*5   +   0   –   (5^2   +   0^2)
25   =   10*5   +   1   –   (5^2   +   1^2)

 
Can a 2-digit number > 25 be expressed in this manner?

 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Math Beauty and tagged . Bookmark the permalink.

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