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Quote:GMMy infinite family gives a solution for every odd number at least 3. The smallest even number I have found was also 52.

Any ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?

If it is A^3 + (A+1)^3 + ... + N^3 = (N+1)^3 + ... + B^3, then this becomes 2 ((N-1) N / 2)^2 = ((A - 1) A / 2)^2 + (B (B+1) / 2)^2, which can be expressed as 2 Q^2 = P^2 + R^2, where P < Q < R and all three are triangular numbers.

The {3, 4, 5, 6} solution has A = 3, N = 5, B = 6; 2 • (5 • 6 / 2)^2 = 450 = 9 + 441 = (2 • 3 / 2)^2 + (6 • 7 / 2)^2; P = 3, Q = 15, and R = 21.

Quote:ThatDonGuyAny ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?

If it is A^3 + (A+1)^3 + ... + N^3 = (N+1)^3 + ... + B^3, then this becomes 2 ((N-1) N / 2)^2 = ((A - 1) A / 2)^2 + (B (B+1) / 2)^2, which can be expressed as 2 Q^2 = P^2 + R^2, where P < Q < R and all three are triangular numbers.

The {3, 4, 5, 6} solution has A = 3, N = 5, B = 6; 2 • (5 • 6 / 2)^2 = 450 = 9 + 441 = (2 • 3 / 2)^2 + (6 • 7 / 2)^2; P = 3, Q = 15, and R = 21.

. . . rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.

Quote:teliotIs there a solution that starts at N = 1?

I am quite confident that there is not. If you assume that 1^2+2^2+...+N^2=(N+1)^2+...+(N+B)^2, then after using the summation formula for squares, you get the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0, which has degree 3 in both B and N. By using the the change of variables x=2(4B+2N+1)/(2B-2N-1), y=-6/(2B-2N-1), this equation can be transformed to the form y^2=x^3-12x+20. (This is not black magic, it is called the Weierstrass form and the curve defined by this equation is an elliptic curve.)

If the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0 has an integer solution for B and N, we get a rational (but not necessarily integer) solution of y^2=x^3-12x+20. The good news is that elliptic curves have been studied extensively and there is a description of the rational solutions of this equation, although it is not as simple as substituting numbers into a formula.

If we know x, y, then B and N are given by B=-(x+2)/(2y), N=(4-x-y)/(2y). y=-6/(2B-2N-1) means that y has to be a (possibly negative) integer dividing 6 or a fraction whose numerator in its lowest terms is a (possibly negative) integer dividing 6.

The integer solutions of y^2=x^3-12x+20 are (x,y)=(-2,6),(4,6),(2,2),(1,3),(10,30),(4,2),(22,102),(89,839) and the same x values with the sign of y changed. None of these give feasible values for B and N. The numerators and denominators of the rational solutions increase quickly, so numerator will never be a divisor 6, but a rigorous proof would require someone who knows more about this topic than I do.

Quote:ThatDonGuyAny ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?

P^2, Q^2, R^2 form an arithmetic progression, the problem of 3 squares forming an arithmetic progression has been solved,

any solution is of the form P/Q=(-t^2+2t+1)/(t^2+1), P/Q=(t^2+2t-1)/(t^2+1), where t is a rational number, if we want P/Q and R/Q to be positive and P/Q<1, then t has to be between 1 and sqrt(2)+1. I am unable post a link, but search for Keith Conrad and arithmetic progression of three squares for a similar formula. (P=3, Q=15, R=21 corresponds to t=2.)

Let t=a/b with a and b coprime, then P/Q=(-a^2+2*a*b+b^2)/(a^2+b^2), R/Q=(a^2+2*a*b-b^2)/(a^2+b^2). a and be cannot both be even as they are coprime. If one of a and b is even and the other is odd, then the numerators and the denominators of these fractions are also coprime, so P=m(-a^2+2*a*b+b^2), Q=m(a^2+b^2), R=m(a^2+2*a*b-b^2) for some integer m. If a and b are both odd, then the greatest common divisor of the the numerators and the denominators is 2, P=m(-a^2+2*a*b+b^2)/2, Q=m(a^2+b^2)/2, R=m(a^2+2*a*b-b^2)/2 for some integer m. Now the question is, can all these numbers be triangular numbers?

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000

Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?

Quote:WizardI don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000

Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?

I assume the winner wins the amount of money he has left. But does does the loser get $0 or some other amount?

Quote:unJonI assume the winner wins the amount of money he has left. But does does the loser get $0 or some other amount?

Runner up gets a flat $2000, but doubt that was contemplated in the question. Need some clarification please.

https://www.youtube.com/watch?v=9R9EhbdxdII

Clearly an all or nothing bid for top score.

9/26/84 Final Jeopardy! - YouTube

https://www.youtube.com/watch?v=O7ck9MWnn6Y

Seems other place finishers kept the money they won.

Super Old School episode:

JEOPARDY! 1974-75 Nighttime Syndicated Season - YouTube

https://www.youtube.com/watch?v=GtyWD9QpFME

If they all won in Final Jeopardy, there would have been a tie. Would they have given out 2 cars and had 2 returning champions?

A 3-Way Tie in Final Jeopardy! | JEOPARDY! - YouTube

https://www.youtube.com/watch?v=KDTxS9_CwZA

The leader could have bet just $600 to win the game, but no, no, no, he didn't do that.

Jeopardy! First: Tiebreaker | JEOPARDY! - YouTube

https://www.youtube.com/watch?v=c7cYON3uVZo

Bonus Daily Double Video Clip:

Norma Tells Norman's Secret | Bates Motel - YouTube

https://www.youtube.com/watch?v=Qes0evIhezI

Very cool that you were able to transform this into an elliptic curve. I know a bit of algebraic geometry, studied it for 2 years in graduate school (40 years ago!). Maybe I could understand some of the more advanced arguments, though surely I am very rusty. Anyway thanks for your superb input here.Quote:GMI am quite confident that there is not. If you assume that 1^2+2^2+...+N^2=(N+1)^2+...+(N+B)^2, then after using the summation formula for squares, you get the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0, which has degree 3 in both B and N. By using the the change of variables x=2(4B+2N+1)/(2B-2N-1), y=-6/(2B-2N-1), this equation can be transformed to the form y^2=x^3-12x+20. (This is not black magic, it is called the Weierstrass form and the curve defined by this equation is an elliptic curve.)

If the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0 has an integer solution for B and N, we get a rational (but not necessarily integer) solution of y^2=x^3-12x+20. The good news is that elliptic curves have been studied extensively and there is a description of the rational solutions of this equation, although it is not as simple as substituting numbers into a formula.

If we know x, y, then B and N are given by B=-(x+2)/(2y), N=(4-x-y)/(2y). y=-6/(2B-2N-1) means that y has to be a (possibly negative) integer dividing 6 or a fraction whose numerator in its lowest terms is a (possibly negative) integer dividing 6.

The integer solutions of y^2=x^3-12x+20 are (x,y)=(-2,6),(4,6),(2,2),(1,3),(10,30),(4,2),(22,102),(89,839) and the same x values with the sign of y changed. None of these give feasible values for B and N. The numerators and denominators of the rational solutions increase quickly, so numerator will never be a divisor 6, but a rigorous proof would require someone who knows more about this topic than I do.