# Problem from GCSE Paired Pilot

Discussion in 'Mathematics' started by GoldMaths, Jun 22, 2011.

1. ### GoldMathsNew commenter

Two square numbers have a sum of 50005.

I have found two solutions:

98^2 + 201^2 = 50005
102^2 + 199^2 = 50005

Question is are there any more?
Is there a logical way to approach this? I have used excel to solve so far....

2. ### GoldMathsNew commenter

54^2 + 217^2 = 50005
87^2 + 206^2 = 50005

3. ### Guzintas

Which paper and which board? Presumably calculator allowed. Seems like T&I at first glance - is that on the syllabus?

4. ### GoldMathsNew commenter

Think it was Methods 2 AQA but cant be 100%.

Ridiculously hard if your supposed to do it using trial and improvement.

5. ### Guzintas

Was there no lead-in part and where was it in the paper ie early on or at the end?

6. ### GoldMathsNew commenter

(a) Show that (mn ? 1)^2
+ (m + n)^2
= (m^2
+ 1)(n^2
+ 1)
[3]

The two shorter sides of a right-angled triangle are 2m and m^2
? 1 where m is a
positive integer.

(b) Show that the hypotenuse is always a positive integer.
[2]

(c) Find two square numbers which have a sum equal to 50005
[2]

7. ### Guzintas

Can do the first two parts but struggling to see the link with part (c) - if there is one.

8. ### bombaysapphireStar commenter

50005 = 5 x 10001
5 = 2^2+1 so m=2
10001 = 100^2+1 so n=100
Square numbers 199 and 102.
So yep - definitely a link from (a) to (c).

10. ### GoldMathsNew commenter

Is there anyway to solve this without a trial and error approach?

11. ### bombaysapphireStar commenter

See post 8. 50005 clearly factorises into 5 and 10001.
These appear to have been chosen because it is easy to spot that they are both 1 bigger than a square number.

13. ### Polecat

If you want to use part (a), then the only starting points are
50005 = 5 X 10001 or 137 X 365 or 73 X 885.
I don't know if there are solutions not using part (a), but I'll tell you later.

14. ### Polecat

Using Maple 13, I got the following complete set of solutions:
(54, 217); (87, 206); (98, 201); (102, 199).
I assume this is what DM found using the Mathematica/Alpha machine.
So yes, I suppose the method is brute force.

15. ### mmmmmathsNew commenter

Not brute force, use part a) see post 8.

16. ### DMNew commenter

Poley knew how to answer the question using the result provided but was answering the wider question about finding the other solutions.

17. ### Guzintas

Nice one. Who said GCSEs were too easy?

18. ### jacksonator

Really nice question - will be using that as a starter in the next week.