You are correct that .
However, there is not any Pythagorean Triad which satisfies this equation. So I would say that this is unsolvable.
Hi
For this question:
The sequences {1, 3, 5, ……,p} and {1, 3, 5, ………, q} contain the integer values of
pand q respectively.
Find the value of p+q if :
{1 + 3 + 5 + ……+ p} +{1 + 3 + 5 + …… + q} = {1 + 3 + 5 + ……+ 33}
I've gotten this far:
0.5 (p+1) [(p+1)/2] + 0.5 (q+1) [(q+1)/2] = 0.5 (33+1) [(33+1)/2]
...
(p+1)^2 + (q+1)^2 = 34^2
What do I do now to get p + q???
Thank you!
You could try...
sum of an arithmetic series is.........
Now we can ask if is the sum of 2 squares.
You can do this by drawing a 17 by 17 square.
Then if we isolate a 16 by 16 square inside it, we are left with 1(17)+1(16)=33
If we isolate a 15 by 15 square, we have 2(17)+2(15)=34+30=64
That's a square, so we have
Hence p (or q) is 29... from
and the other (in this case q) is
Hence p+q=44 is a solution, for p=29 and q=15.
However, that will not hold for other values of p+q that give 44 obviously,
since clearly p or q cannot be >33 among other things.
you can try a few more combinations to see if there are other solutions
etc
crikey!! Krahl got there first, was i really that long typing this??