## sum of the squares of primes equalling a square

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• Some tidbits that some may find interesting and perhaps fun to prove: The sum of the squares of 3 primes never equals a square. If the sum of the squares of 5
Message 1 of 7 , Jul 6, 2012
Some tidbits that some may find interesting and perhaps fun to prove:

The sum of the squares of 3 primes never equals a square.

If the sum of the squares of 5 primes equals a square, two of the primes are 2 and 3.

(For example 2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 16^2)

If the sum of the squares of 7 primes equals a square, three of the primes are 2.

(For example 2^2 + 2^2 + 2^2 + 3^2 + 5^2 + 13^2 + 19^2 = 24^2 )

Mark
• ... 8^2 + 49^2 + 64^2 = 81^2 For any integers p, q, s, and t let d=(p^2+q^2+s^2+t^2)/2 c=(p^2+q^2-s^2-t^2)/2 a=ps+qt b=pt-qs then a^2 + b^2 + c^2 = d^2
Message 2 of 7 , Jul 6, 2012
--- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@...> wrote:
>
> Some tidbits that some may find interesting and perhaps fun to prove:
>
> The sum of the squares of 3 primes never equals a square.

8^2 + 49^2 + 64^2 = 81^2

For any integers p, q, s, and t let

d=(p^2+q^2+s^2+t^2)/2
c=(p^2+q^2-s^2-t^2)/2
a=ps+qt
b=pt-qs

then a^2 + b^2 + c^2 = d^2

Depending on the choices, c and d might be halves, and all the values will need be doubled to get integers.

William Lipp
Poohbah of OddPerfect.org
• Yes, the sum of squares of 3 composites can equal a square. But Mark stated that the sum of squares of 3 primes cannot; it s pretty easy to prove. (Choose the
Message 3 of 7 , Jul 6, 2012
Yes, the sum of squares of 3 composites can equal a square.

But Mark stated that the sum of squares of 3 primes cannot;
it's pretty easy to prove. (Choose the right modulus for
solving your Diophantine equation and it becomes clear.)

On 7/6/2012 8:53 AM, elevensmooth wrote:
>
>
> --- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@...> wrote:
>>
>> Some tidbits that some may find interesting and perhaps fun to prove:
>>
>> The sum of the squares of 3 primes never equals a square.
>
> 8^2 + 49^2 + 64^2 = 81^2
>
> For any integers p, q, s, and t let
>
> d=(p^2+q^2+s^2+t^2)/2
> c=(p^2+q^2-s^2-t^2)/2
> a=ps+qt
> b=pt-qs
>
> then a^2 + b^2 + c^2 = d^2
>
> Depending on the choices, c and d might be halves, and all the values will need be doubled to get integers.
>
> William Lipp
> Poohbah of OddPerfect.org
>
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://primes.utm.edu/
>
>
>
>
>
>
• True. And those identities from William were pretty neat. I had said, If the sum of the squares of 5 primes equals a square, two of the primes are 2 and 3
Message 4 of 7 , Jul 6, 2012
True. And those identities from William were pretty neat.

"If the sum of the squares of 5 primes equals a square, two of the primes are 2
and 3"

which was a little vague. It should really have been

If the sum of the squares of 5 primes equals a square, one of the primes is 2 and another is 3.

Besides the previous

If the sum of the squares of 7 primes equals a square, three of the primes are
2.

we now have this:

If the sum of the squares of 11 primes equals a square, two of the primes are 2 and another is 3.

I never would have guessed off the top of my head that with this many squares there would be such restrictions!

Mark

--- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
>
> Yes, the sum of squares of 3 composites can equal a square.
>
> But Mark stated that the sum of squares of 3 primes cannot;
> it's pretty easy to prove. (Choose the right modulus for
> solving your Diophantine equation and it becomes clear.)
>
> On 7/6/2012 8:53 AM, elevensmooth wrote:
> >
> >
> > --- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@> wrote:
> >>
> >> Some tidbits that some may find interesting and perhaps fun to prove:
> >>
> >> The sum of the squares of 3 primes never equals a square.
> >
> > 8^2 + 49^2 + 64^2 = 81^2
> >
> > For any integers p, q, s, and t let
> >
> > d=(p^2+q^2+s^2+t^2)/2
> > c=(p^2+q^2-s^2-t^2)/2
> > a=ps+qt
> > b=pt-qs
> >
> > then a^2 + b^2 + c^2 = d^2
> >
> > Depending on the choices, c and d might be halves, and all the values will need be doubled to get integers.
> >
> > William Lipp
> > Poohbah of OddPerfect.org
> >
> >
> >
> >
> > ------------------------------------
> >
> > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> > The Prime Pages : http://primes.utm.edu/
> >
> >
> >
> >
> >
> >
>
• ... Most likely attack is mod 24. Squares are {0,1,4,9,16,12}. Prime squares are {1} and the trivial (choice-free) {4,9} So we have some 2s, some 3s, and then
Message 5 of 7 , Jul 6, 2012
--- On Fri, 7/6/12, Mark <mark.underwood@...> wrote:
> Some tidbits that some may find interesting and perhaps fun to prove:
>
> The sum of the squares of 3 primes never equals a square.
>
> If the sum of the squares of 5 primes equals a square, two
> of the primes are 2 and 3.
>
> (For example 2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 16^2)

Most likely attack is mod 24.
Squares are {0,1,4,9,16,12}.
Prime squares are {1} and the trivial (choice-free) {4,9}

So we have some 2s, some 3s, and then some things that square to 1:

\ 0*2 1*2 2*2 3*2 4*2 5*2
0*3 0+5 4+4 8+3 12+2 16+1 20+0 | no solns
1*3 9+4 13+3 17+2 21+1 1+0 x | 13+3 and 1+0 are squares
2*3 18+3 22+2 2+1 6+0 x x | 22+2 is a square
3*3 3+2 7+1 11+0 x x x | no solns
4*3 12+1 16+0 x x x x | 16+0 is a square
5*3 21+0 x x x x x | no solns

13+3 is {2, 3, p, q, r}
1+0 is {2, 2, 2, 2, 3}
22+2 is {2, 3, 3, p, q}
16+0 is {2, 3, 3, 3, 3}

> If the sum of the squares of 7 primes equals a square, three
> of the primes are 2.

Presuming the above attack works too:

\ 0*2 1*2 2*2 3*2 4*2 5*2 6*2 7*2
0*3 0+7 4+6 8+5 12+4 16+3 20+2 0+2 4+2 | 12+4 has 3*2
1*3 9+6 13+5 17+4 21+3 1+2 5+1 9+0 x | 21+3 and 9+0 have >=3*2
2*3 18+5 22+4 2+3 6+2 10+1 14+0 x x | no solns
3*3 3+4 7+3 11+2 15+1 19+0 x x x | 15+1 has 3*2
4*3 12+3 16+2 20+1 0+0 x x x x | 0+0 has 3*2
5*3 21+2 1+1 5+0 x x x x x | no solns
6*3 6+1 10+0 x x x x x x | no solns
7*3 15+0 x x x x x x x | no solns

Confirmed.

Whenever there's a puzzle involving squares, 24 is a very powerful lever.

Phil
• ... Phil wisely chose to use 24 = 2^3 * 3 for a modular exhaustion. When I first encountered number theory, it took me a while to understand why working mod 8
Message 6 of 7 , Jul 6, 2012
Phil Carmody <thefatphil@...> wrote:

> Most likely attack is mod 24.
....

Phil wisely chose to use 24 = 2^3 * 3
for a modular exhaustion.

When I first encountered number theory,
it took me a while to understand why
working mod 8 was important, when it seemed
to me (wrongly) that we need only care

If one will glance, for example, at
http://en.wikipedia.org/wiki/Kronecker_symbol
then one will see that mod 8 criteria really matter.
[Also when trying to extract modular square roots.]

Silly summary: when dealing with any even prime,
it may be wise to work modulo its cube.

David
• ... Thank you Phil, that was very helpful. Now if it could only level a house made from hypothetical perfect euler bricks! (That happens to be the problem
Message 7 of 7 , Jul 7, 2012
> Whenever there's a puzzle involving squares, 24 is a very powerful lever.

Thank you Phil, that was very helpful. Now if it could only level a house made from hypothetical perfect euler bricks! (That happens to be the problem that led me down this particular sum of squares of primes theme, albeit as a side diversion!)

Mark

--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
>
> --- On Fri, 7/6/12, Mark <mark.underwood@...> wrote:
> > Some tidbits that some may find interesting and perhaps fun to prove:
> >
> > The sum of the squares of 3 primes never equals a square.
> >
> > If the sum of the squares of 5 primes equals a square, two
> > of the primes are 2 and 3.
> >
> > (For example 2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 16^2)
>
> Most likely attack is mod 24.
> Squares are {0,1,4,9,16,12}.
> Prime squares are {1} and the trivial (choice-free) {4,9}
>
> So we have some 2s, some 3s, and then some things that square to 1:
>
> \ 0*2 1*2 2*2 3*2 4*2 5*2
> 0*3 0+5 4+4 8+3 12+2 16+1 20+0 | no solns
> 1*3 9+4 13+3 17+2 21+1 1+0 x | 13+3 and 1+0 are squares
> 2*3 18+3 22+2 2+1 6+0 x x | 22+2 is a square
> 3*3 3+2 7+1 11+0 x x x | no solns
> 4*3 12+1 16+0 x x x x | 16+0 is a square
> 5*3 21+0 x x x x x | no solns
>
> 13+3 is {2, 3, p, q, r}
> 1+0 is {2, 2, 2, 2, 3}
> 22+2 is {2, 3, 3, p, q}
> 16+0 is {2, 3, 3, 3, 3}
>
>
> > If the sum of the squares of 7 primes equals a square, three
> > of the primes are 2.
>
> Presuming the above attack works too:
>
>
> \ 0*2 1*2 2*2 3*2 4*2 5*2 6*2 7*2
> 0*3 0+7 4+6 8+5 12+4 16+3 20+2 0+2 4+2 | 12+4 has 3*2
> 1*3 9+6 13+5 17+4 21+3 1+2 5+1 9+0 x | 21+3 and 9+0 have >=3*2
> 2*3 18+5 22+4 2+3 6+2 10+1 14+0 x x | no solns
> 3*3 3+4 7+3 11+2 15+1 19+0 x x x | 15+1 has 3*2
> 4*3 12+3 16+2 20+1 0+0 x x x x | 0+0 has 3*2
> 5*3 21+2 1+1 5+0 x x x x x | no solns
> 6*3 6+1 10+0 x x x x x x | no solns
> 7*3 15+0 x x x x x x x | no solns
>
> Confirmed.
>
> Whenever there's a puzzle involving squares, 24 is a very powerful lever.
>
> Phil
>
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