Part a) is a typical problem for induction. If you have 2 lines satisfiing the Fibonacci condition with numbers b_0,b_1,... and c_0,c_1,... then you have in the next line you have numbers of the form b_i+c_(i-1). So you have to prove: b_i+c_(i-1)+b_(i+1)+c_i=b_(i+2)+c_(i+1) for all i. That follows because of the Fibonacci structure of the previous lines, q.e.d. Now, I try to solve b) but it seems to be more difficult!

Actually, even part b) is not really difficult... I call "horizontal construction law" the law that every number equals the sum of its two precedents in its row (this was established in part a) of the problem), and I call "vertical construction law" the law that every number equals the sum of its two precedents in its column (this was given in the problem). If you consider just the four consecutive lines starting with a, b, c, d, then you get the following partial array: a ? ? ? ... b ? ? ... c ? ... d ... Now, if we call u the second number in the first line, then we can fill up the first line by the horizontal construction law: a u a+u a+2u ... b ? ? ... c ? ... d ... Now, by the vertical construction law, it follows that (a+u) + (the second number in the second row) = c, hence the second number in the second row is c - (a+u). Thus the partial table takes the form a u a+u a+2u ... b c-(a+u) ? ... c ? ... d ... Now, by the horizontal construction law in the second line, this becomes a u a+u a+2u ... b c-(a+u) b+(c-(a+u)) ... c ? ... d ... Now, if we call T the remaining second element of the third line: a u a+u a+2u ... b c-(a+u) b+(c-(a+u)) ... c T ... d ... then the vertical construction law yields both (a+2u) + (b+(c-(a+u))) = T and (b+(c-(a+u))) + T = d. Thus, d - (b+(c-(a+u))) = T = (a+2u) + (b+(c-(a+u))). After some simplification, this becomes d + a = 2b + 2c, and thus d = 2b + 2c - a. This is what we wanted, if I have not made some hundreds of mistakes on the way. And even if it's right, I guess this solution will be rated the worst-written solution of all MathLinks; yes, I'm sorry, I'm very tired now and besides, I'm not used to write up such solutions (and before Valentin complains about my rudimentary use of [ code ] tags instead of LaTeX, I must say I haven't succeeded to write the arrays with LaTeX without ending up with [unparseable and potentially dangerous formula]s all the time). Puh, it's time to go to bed. Darij

Darij, your solution is really true, I've checked it now. Good night, I go to bed too!

Misha, thank you for the verification. One final thing I would like to note: If one wouldn't restrict oneself on the "upper right" part of the table only, and one would consider the whole table, placing a 1 on the leftmost place of the second line, then one would get the following "completed" table: 0 1 1 2 3 5 8 1 0 1 1 2 3 5 1 1 2 3 5 8 13 2 1 3 4 7 11 18 3 2 5 7 12 19 31 5 3 8 11 19 30 49 8 5 13 18 31 49 80 I hope I've made no calculation mistakes here, but anyway the moral is clear: The table is symmetric with respect to its main diagonal, i. e. the n-th number of the m-th row equals the m-th number of the n-th row. But this is really a kind of trivial. Darij