(My failed attempt to solve this is below.) ----- Here's a (somewhat late) Diophantine puzzle in honor of the 365th day of the year: It is not hard to show that 365 can be written as both 10^2 + 11^2 + 12^2 and 13^2 + 14^2. (This fact came to my attention about twenty years ago through Nikolai Petrovich Bogdanov-Belsky's painting "Mental Calculation in the Public School Of S. A. Rachinsky", which you can view at https://goo.gl/images/yo88LY. Puzzle: Show that there are infinitely many integers that can be written both as a sum of two consecutive squares and as a sum of three consecutive squares. For extra credit, find the proof mentally. :-) ----- Let's see: The sum of 3 consecutive squares is (x-1)^2 + x^2 + (x+1)^2 = 3x^2 + 2, x in Z while for 2 consecutive squares, y^2 + (y+1)^2 = 2y^2 + 2y + 1, y in Z. Equality means 2y^2 + 2y + 1 = 3x^2 + 2 or 2y^2 + 2y = 3x^2 + 1 and the LHS is even, hence the RHS is even, so x is odd: say x = (2K + 1). So 2y(y+1) = 3(2K+1)^2 + 1 = 12K^2 + 12K + 2 or y(y+1) = 6K^2 + 6K + 1. Now the LHS must be even but the RHS must be odd, so this is impossible. (Arggh, probably made some dumb mistake.) —Dan
On 12/31/2017 6:49 PM, Dan Asimov wrote:
(My failed attempt to solve this is below.)
----- Here's a (somewhat late) Diophantine puzzle in honor of the 365th day of the year:
It is not hard to show that 365 can be written as both 10^2 + 11^2 + 12^2 and 13^2 + 14^2. (This fact came to my attention about twenty years ago through Nikolai Petrovich Bogdanov-Belsky's painting "Mental Calculation in the Public School Of S. A. Rachinsky", which you can view at https://goo.gl/images/yo88LY.
Puzzle: Show that there are infinitely many integers that can be written both as a sum of two consecutive squares and as a sum of three consecutive squares.
For extra credit, find the proof mentally. :-) -----
Let's see: The sum of 3 consecutive squares is
(x-1)^2 + x^2 + (x+1)^2 = 3x^2 + 2, x in Z
while for 2 consecutive squares,
y^2 + (y+1)^2 = 2y^2 + 2y + 1, y in Z.
Equality means
2y^2 + 2y + 1 = 3x^2 + 2
or 2y^2 + 2y = 3x^2 + 1
and the LHS is even, hence the RHS is even, so x is odd: say x = (2K + 1). So
2y(y+1) = 3(2K+1)^2 + 1
= 12K^2 + 12K + 2
Should be 12K^2 + 12K + 4 Brent
or y(y+1) = 6K^2 + 6K + 1.
Now the LHS must be even but the RHS must be odd, so this is impossible.
(Arggh, probably made some dumb mistake.)
—Dan
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I gave the problem to Julian. After seven minutes of staring into space he said... Pell’s equation. That right? Regards, Jon
On Dec 31, 2017, at 7:58 PM, Brent Meeker <meekerdb@verizon.net> wrote:
On 12/31/2017 6:49 PM, Dan Asimov wrote: (My failed attempt to solve this is below.)
----- Here's a (somewhat late) Diophantine puzzle in honor of the 365th day of the year:
It is not hard to show that 365 can be written as both 10^2 + 11^2 + 12^2 and 13^2 + 14^2. (This fact came to my attention about twenty years ago through Nikolai Petrovich Bogdanov-Belsky's painting "Mental Calculation in the Public School Of S. A. Rachinsky", which you can view at https://goo.gl/images/yo88LY.
Puzzle: Show that there are infinitely many integers that can be written both as a sum of two consecutive squares and as a sum of three consecutive squares.
For extra credit, find the proof mentally. :-) -----
Let's see: The sum of 3 consecutive squares is
(x-1)^2 + x^2 + (x+1)^2 = 3x^2 + 2, x in Z
while for 2 consecutive squares,
y^2 + (y+1)^2 = 2y^2 + 2y + 1, y in Z.
Equality means
2y^2 + 2y + 1 = 3x^2 + 2
or 2y^2 + 2y = 3x^2 + 1
and the LHS is even, hence the RHS is even, so x is odd: say x = (2K + 1). So
2y(y+1) = 3(2K+1)^2 + 1
= 12K^2 + 12K + 2
Should be 12K^2 + 12K + 4
Brent
or y(y+1) = 6K^2 + 6K + 1.
Now the LHS must be even but the RHS must be odd, so this is impossible.
(Arggh, probably made some dumb mistake.)
—Dan
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participants (3)
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Brent Meeker -
Dan Asimov -
Jon Ziegler