Turning to cubic diophantine equations, these are notoriously difficult to solve but there has been enormous progress in recent years with elliptic curves. This paper looks promising since I believe our cubic can be converted into a solvable form and we already have one rational point to kick off the analysis. But it is hard work and I am not sure that I will have the time that understanding this paper demands.

]]>I strongly approve of your efforts to (a) prove the uniqueness of solutions and (b) correct false statements when they appear.

It has effectively been claimed (without proof) on your other site that

(m, n) = (17, 42) is the only solution in positive integers of the Diophantine equation

43m(m+1)(2m+1) = 3n(n+1)(2n+1).

I would like to know whether the claim is true.

Tony

]]>\[S(2m)=(115/100)\left\{S(2m)-S(m)\right\}\Rightarrow 3\,S(2m)=23\,0S(m)\]

On expanding the sums of squares, this becomes

\[3\,(2m)(2m+1)(4m+1)/6=23\,m(m+1)(2m+1)/6\]which can be simplified to give:\[m(2m+1)(m-17) = 0\]Since we are only interested in positive values for m, this gives \(m = 17\).

The second condition is:\[S(n)=(215/200)\left\{S(n)-S(m)\right\}\Rightarrow 3\,S(n)=43\,S(m)\]

Expanding and substituting for \(m\) now gives \[n(n+1)(2n+1)=2\times3\times5\times7\times17\times43\] Here \(2n^3<153510\), which means that \(n < 43\), and one of the terms on the left hand side must have 43 as a factor since 43 on the right hand side is a prime. Hence either \((n + 1) = 43\) or \((2n+1)=43\); substituting these values into the equation shows that only \(n=42\) provides a solution so the above equation simplifies to: \[(n-42)(2\,n^2+ 87\,n+3655)\;=\;0\]where the quadratic term might give two additional solutions. In fact it doesn't, since it has two complex conjugate roots: \(n=(1/4)(-87\pm\sqrt{21671}\,i)\) and this leaves \(n=42\) as the only real solution. The answer is hence a final payment of £1764.

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from itertools import count S = lambda n: n * (n + 1) * (2 * n + 1) // 6 for m in count(1): sm, s2m = S(m), S(2 * m) if 100 * s2m == 115 * (s2m - sm): for n in count(m): sn = S(n) if 200 * sn == 215 * (sn - sm): print(f"£{n ** 2} (birthdays {m + 1}..{n})") exit() |