- Welcome

I have writed with this problem recently, but I still can not solve it.

I need to solve a equation:

a^2+b^2=c^2

where a and c are primes and b is a product of not more then four prime

numbers. I have found that all solutions are:

1) a=3 b=4=2*2 c=5

2) a=5 b=12=2*2*3 c=13

3) a=11 b=60=2*2*3*5 c=61

I have also concluded that:

a=2q+1

b=2q^2+2q

c=2q^2+2q+1

and:

b=(a^2-1)/2

c=(a^2+1)/2

But I still don't know how should I proof that these three solutions

are the only one. How should I do proof it?

I would be grateful for any help.

Best Regards

Adam Karasek

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[Non-text portions of this message have been removed] - --- In primenumbers@yahoogroups.com, Adam Karasek

<adam_karasek@y...> wrote:> Welcome

still can not solve it.

>

> I have writed with this problem recently, but I

> I need to solve a equation:

more then four prime

> a^2+b^2=c^2

> where a and c are primes and b is a product of not

> numbers. I have found that all solutions are:

three solutions

> 1) a=3 b=4=2*2 c=5

> 2) a=5 b=12=2*2*3 c=13

> 3) a=11 b=60=2*2*3*5 c=61

> I have also concluded that:

> a=2q+1

> b=2q^2+2q

> c=2q^2+2q+1

> and:

> b=(a^2-1)/2

> c=(a^2+1)/2

> But I still don't know how should I proof that these

> are the only one. How should I do proof it?

Indeed these 3 solutions you found are all solutions

>

> I would be grateful for any help.

> Best Regards

> Adam Karasek

for the equation.

1) a=3 b=4=2*2 c=5

2) a=5 b=12=2*2*3 c=13

3) a=11 b=60=2*2*3*5 c=61

Your conclusions are also right.

For any a < 13 prime there is no other solution.

If a >= 13 prime:

a=p , p-odd prime p>=13.

b=(p^2-1)/2 , no more than 4 prime factors(might be

equally few)

c=(p^2+1)/2 , also odd prime

Generally for p-odd prime:

p mod10 = 1 or 3 or 7 or 9 , then

p^2 mod10 = 1 or 9 .

1). p^2 mod10 = 9 then

c mod10 = 5 , c-odd prime , c > 5 and 5|c,

contradiction

no solution in this case.

2). p^2 mod10 = 1 then

b mod10 = 0

so 10|b then 5|b

b=2*((p-1)/2)*((p+1)/2),

(p-1)/2,(p+1)/2 are 2 consecutive integers, then their

product is divisible by 2, then (2*2)|b

(p-1)/2,(p+1)/2, (p+3)/2 are 3 consecutive integers,

then their product divisible by 3.

(p+3)/2 is not divisible by 3, if it is, then p+3 is

divisible by 6 p divisible by 3 also p prime

contradiction

((p-1)/2)*((p+1)/2) divided by 3, then 3|b

so (2*2*3*5)|b already 4 prime factors dividing b,

60|b.

but p>=13 then b=(p^2-1)/2 >= 84 > 60.

there is at least one other prime factor dividing b

and (b/60)

so if p^2 mod10 = 1 , b is at least product of 5 prime

factors

does not correspond to initial conditions.

no solution in this case also

The problem is proved. (seems looking "known" to me)

I hope this will help you.

Best regards,

Dan Dima

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