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PEN Math problems from the collection "Problems in Elementary Number Theory"

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PEN Math problems from the collection "Problems in Elementary Number Theory"

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Peter

#1 h May 25, 2007, 12:26 AM • 2 Y

Y by Adventure10, Mango247

Is there a $3 \times 3$ magic square consisting of distinct Fibonacci numbers (both $f_{1}$ and $f_{2}$ may be used; thus two $1$ s are allowed)?

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469 posts

bboypa

#2 h Mar 6, 2008, 12:30 AM • 2 Y

Y by Adventure10, Mango247

no, it doesn't exist.

we have:

A B C
D E F
G H I.

$f_n$ is the greatest number that we use.

CASE 1) $f_n$ is in A (or simmetrically in C,G,I)
we must have A+B+C=C+F+I so A+B=C+I but C+I $\le$ A < A+B, absurd.

CASE 2) $f_n$ is in B(or simmetrically in D,F,H)
we must have B+E+H=D+E+F so B+H=D+F but D+F $\le$ B < B+H, absurd.

LAST CASE) $f_n$ is in E.
we must have G+H+I=A+B+C=D+E+F>E. but $\frac{(G+H+I)+(A+B+C)}{2} \le \frac{\sum_{i=n-6}^{n-1}{f_i}}{2}=\frac{(f_{n-1}+f_{n-2})+(\sum_{n-6}^{n-3}{f_i})}{2}< f_n$ , absurde

bye
paolo

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Woett

#3 h May 11, 2010, 4:40 PM • 2 Y

Y by Adventure10, Mango247

an easier approach in which we prove that even the sum of 3 different rows cannot be the same;

Let f(n+4) be the largest occuring Fibonacci number. Then there must be a row in which the largest Fibonacci number is no larger than
f(n+2). But the sum of that row is no larger than f(n) + f(n+1) + f(n+2) = f(n) + f(n+3) < f(n+2) + f(n+3) = f(n+4), impossible.

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Woett

#4 h May 13, 2010, 6:10 PM • 1 Y

Y by Adventure10

Note that the previous proof alsof easily generalizes to m*m magic squares; If we let f(m+n-1) be the largest occuring Fibonacci number, there must be a row in which the largest Fibonacci number is no larger than f(n). The sum in that row is smaller than or equal to: f(n-m+1) + f(n-m+2) + .. + f(n) < (m-1)f(n) < 1.5^(m-1)f(n) < f(n+m-1), contradiction.

Another approach; Let X = F(x) = the x-th fibonacci number. Let

A B C
D E F
G H I

be our 3*3 magic square. Then A + I = 2E = C + G. This implies that max(A, C, G, I) is at least f(e+2). But this is already larger than 2E, contradiction.

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