Explain your p&p method of finding two integers, their squares sum being 685.
You may not use algebraic formulae, calculator or Google.
Be D3 wise to make it D1!
No sure what to make of the "D3... D1.." comment, and maybe my way is, well, uninspired, but here is what I did (p&p, which took a few min.s):
What pairs of last digits squared yield last digits summing to 5? Ans.: (0,5),
(1,2), (1,8), (9,2), (9,8), (3,4), (3,6), (7,4), (7,6)
(0,5) is not it, since 685-10^2 is not a square.
Compile some squares, while keeping them under 685. E.g., for (1): 1^2, 11^2, 21^2 = (1,121,441)
(2) = (4,144,484) so (1,2) ain't it.
(8) = (64, 324) , so (1,8) ain't it.
(9) = (81,361)
Ah, (9,8) works: (19^2 + 18^2)= (361 + 324) = 685.
There may be more pairs, but, having seen the way forward, I quit.
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Edited on November 29, 2018, 1:20 pm
soln
