Pentagonal, triangular, octagonal numbers

**Denny Crane** · July 13th 2009, 01:14 AM

If we have a formula Un+Pn+Tn for n, i.e. Un is the nth term of the octagonal number set etc. Is it possible for Un+Pn+Tn=166?

I tried doing this by equating the formulas, but then I get a quadratic and have to complete the square and I don't think it is right. Anyone got any ideas?

e.g:
Un = n(3n-2)
Tn = (n^2 + n)/2
Pn = n(3n-1)/2

**HallsofIvy** · July 13th 2009, 08:04 AM

Originally Posted by Denny Crane

If we have a formula Un+Pn+Tn for n, i.e. Un is the nth term of the octagonal number set etc. Is it possible for Un+Pn+Tn=166?

I tried doing this by equating the formulas, but then I get a quadratic and have to complete the square and I don't think it is right. Anyone got any ideas?

e.g:
Un = n(3n-2)
Tn = (n^2 + n)/2
Pn = n(3n-1)/2

Okay, Un+ Tn+ Pn= $n(3n-2)+ \frac{n^2+ n}{2}+ \frac{n(3n-1)}{2}$ $= \frac{6n^2- 4n}{2}+ \frac{n^2+ n}{2}+ \frac{3n^2- n}{2}$ $= \frac{6n^2- 4n+ n^2+ n+ 3n^2- n}{2}$ $= \frac{10n^2- 3n}{2}= 166$ so the quadratic equation you get is just $10n^2- 3n= 332$ . Yes, completing the square is "non-trivial" but it is possible. I would first divide the equation by 10 to get $n^2- .3n= 33.2$ . Half of -.3 is -.15 and its square is 0.0225. $n^2- .3n+ 0.0225= (n- .15)^2= 33.2+ 0.0225= 33.2225$ so $n= .15+ \sqrt{33.225}$ . Now, all you need to do is observe that that result is NOT a positive integer.

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