## Equation with primes

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• 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
Message 1 of 2 , Nov 1, 2003
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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

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• ... still can not solve it. ... more then four prime ... three solutions ... Indeed these 3 solutions you found are all solutions for the equation. 1) a=3
Message 2 of 2 , Nov 18, 2003
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> 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

Indeed these 3 solutions you found are all solutions
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
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,
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
((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)