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# Number Thoery Problem watch

1. let [a,b] denote a pair of positive integers such that
a^2 + b^2 = square number

prove that if a is prime, there is a unique corresponding value for b.
2. (Original post by fishpaste)
square numbers only have three factors, 1, their squareroot, and themself, apart from 1, which has 1. ( * )
Bugger, just realised that isn't remotely true. Oh well. Sorry.

However, it's only not true for squares of squares, perhaps you could adapt it or something.
3. If you instead break it down as

a^2 = c^2 - b^2

and use the fact that

"a^2 has factors 1, a and a^2 only" - which is true as a is prime

then you can uniquely work out b and c from a much as you did, remembering that a,b,c are positive.
4. (Original post by RichE)
If you instead break it down as

a^2 = c^2 - b^2

and use the fact that

"a^2 has factors 1, a and a^2 only" - which is true as a is prime

then you can uniquely work out b and c from a much as you did, remembering that a,b,c are positive.
a^2 = c^2 - b^2

a^2 = (c + b)(c - b)

since a is prime, a^2 has factors only 1,a,a^2

the factors can't be (a)(a) because b is a positive integer, so they must be 1 and a^2 respectively

so:
c - b = 1

this alone says that as long as you have c, you can find b, since b = c - 1
QED
(and a^2 = 2c - 1 = 2b + 1)

like this?
5. Well done, now try this extension:

let [a,b] denote a pair of positive integers such that
a^2 + b^2 = square number

prove that there are as many different values for b as there are proper factors of a (Hence proving that prime numbers a have only one corresponding value for b, having only one proper factor)
6. (Original post by Tomp11)
Well done, now try this extension:

let [a,b] denote a pair of positive integers such that
a^2 + b^2 = square number

prove that there are as many different values for b as there are proper factors of a (Hence proving that prime numbers a have only one corresponding value for b, having only one proper factor)
Same again, let the square number be c

=> a^2 = (c-b)(c+b)

Now it follows that c-b can be 1,a or any proper factor of a, and so, since c is fixed, the number of possibilities for b is determined by the number of proper factors of a.
7. (Original post by beauford)
Same again, let the square number be c

=> a^2 = (c-b)(c+b)

Now it follows that c-b can be 1,a or any proper factor of a, and so, since c is fixed, the number of possibilities for b is determined by the number of proper factors of a.
I'm not sure in what sense c is fixed; for example,

8^2 + 15^2 = 17^2 and 8^2 + 6^2 = 10 ^2 [Two different cs]

But more importantly, I don't believe the claim.

I guess you have to count 1 (or a) as a proper factor (otherwise the original problem when a is prime is wrong). In which case 8 has three proper factors 1,2,4 but there are two solutions for b.

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