Given the positive integers, x, y, and z, are consecutive terms of an** arithmetic progression**, **the least value of the positive integer, n,** for which the equation, x^2 -y^2- z^2 = n, has exactly two solutions is n = 27:

34^2- 27^2- 20^2 = 12^2- 9^2-6^2 = 27

It turns out that n = 1155 is the least value which has exactly ten solutions.

How many values of n less than one million have exactly ten distinct solutions?

I'm having problem with counting N:

let's say we're thinking about arithmetic progression that progresses by 10. since 0<n<1,000,000, here are all the possible cases:

Code: Select all

```
P|||| X Y Z N
p==10 21 11 1 319
p==10 22 12 2 336
p==10 23 13 3 351
p==10 24 14 4 364
p==10 25 15 5 375
p==10 26 16 6 384
p==10 27 17 7 391
p==10 28 18 8 396
p==10 29 19 9 399
p==10 30 20 10 400
p==10 31 21 11 399
p==10 32 22 12 396
p==10 33 23 13 391
p==10 34 24 14 384
p==10 35 25 15 375
p==10 36 26 16 364
p==10 37 27 17 351
p==10 38 28 18 336
p==10 39 29 19 319
p==10 40 30 20 300
p==10 41 31 21 279
p==10 42 32 22 256
p==10 43 33 23 231
p==10 44 34 24 204
p==10 45 35 25 175
p==10 46 36 26 144
p==10 47 37 27 111
p==10 48 38 28 76
p==10 49 39 29 39
```

now the problem is I'm having hard time understanding what

**n** I should count.

**the least value of the positive integer, n,**
I can see that 49, 39, 29 yields least value of n, which is 39.

(and this is trivial without enumerating all x, y, z)

initially, given this observation, I thought what I had to do was:

for p==10, the least value of n is 39, count it once, move along to next p

however, by this method the given example of n=1155 having 10 solutions could not be found.

edit :

disregard everything I wrote.

however I have a question : how could n have a 'least' value? doesn't that term implies there is also a 'most' value and there are more than one possible 'n's for an arithmetic sequence?