Frog Frolics (Posted on 2011-11-15)

	Submitted by broll
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First question: Freddie can never reach the shore safely. The numbers 2,10,18... correspond to successive values of (8k+2), while the numbers (8,24,48...) correspond to succesive odd squares. Hence the problem as set is equivalent to proving that every number of the form (8k+3) is the sum of 3 odd squares, a problem better-known as the theorem that 'every number is the sum of no more than 3 triangular numbers', first attributed to Fermat: 1. Let Z be any whole number; the claim is that Z=a+b+c, where a=(x-1)(x)/2, b=(y-1)(y)/2, c=(z-1)(z)/2. I Since 2Z=2(a+b+c), by substitution, 2Z=x(x-1)+y(y-1)+z(z-1) II We complete squares to obtain: 2Z=1/4(2x-1)^2-1/4+1/4(2y-1)^2-1/4+1/4(2z-1)^2-1/4 III Multiply out to dispose of the fractions: 8Z+3=(2x-1)^2+(2y-1)^2+(2z-1)^2: in the problem the last of these is just one less than the square so we have the 3 odd squares and the intervals of (8k+2) as stipulated. 2. Hence another way of putting the same claim is that every number that is 3 more than a multiple of 8 can be expressed as a sum of not more than 3 odd squares. Since 8Z+3<>4^k(8k+7), we can then simply use Fermat's result on sums of squares to confirm that the equality always holds. Second question: Even though he can never reach the shore, Freddie can make any number of hops; provided we are prepared to abandon the condition that Freddie must reach the shore, we can always insert a multiple of a fresh (4m+3) prime into every potential safe haven below a given length. The numbers quickly become large though!

	Subject	Author	Date
	draft solution for comment	broll	2011-11-20 07:34:15
	re(2): Some more thoughts and more questions	broll	2011-11-16 23:06:37
	re: Some more thoughts and more questions	Justin	2011-11-16 20:41:03
	Some more thoughts and more questions	Jer	2011-11-16 09:07:08
	re: Some thoughts and questions	broll	2011-11-15 22:56:23
	Some thoughts and questions	Jer	2011-11-15 16:07:16