How many
integer values of n satisfy this equation?
(6+n)n + (7+n)n = 12n + 1
Prove that there are no others.
I wonder of this problem was devised because of the Taxicab Number story where Ramanujan tells Hardy the number 1729 is interesting because it is the sum of two cubes in two different ways.
For the problem at hand, when n=3 we have 9^3+10^3=1729=12^3+1^3.
As for other solutions n=0 is obvious and n=2 gives 8^2+9^2=12^1+1
n=1 does not work nor does n=4. For n=5 or above the RHS is clearly greater.
On the negative side there are no solutions at -1,-2,-3,-4,-5 or -8 and the RHS is undefined at n=-6 and -7. For n<-8 the RHS approaches zero (when defined) whereas the LHS approaches 1.
There are two nearly integer solutions:
n=-5 gives 1+1/2^5 and 1+1/12^5. The real solution is closer to -4.99397
n=-8 gives 1+1/2^8 and 1+1/12^8. The real solution may be approximated by a rational number a bit below this around -75004/9375
So the exact integer solutions x={0,2,3}
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Posted by Jer
on 2021-11-22 09:32:14 |