[seqfan] Re: Is this possible?

[Andrew Weimholt]

On Sat, Jun 5, 2010 at 2:12 PM, Andrew Weimholt
<andrew.weimholt at gmail.com> wrote:
> On Sat, Jun 5, 2010 at 8:35 AM, Hans Havermann <pxp at rogers.com> wrote:
>> Quadruples {a,b,c,d}, a>b>c>d>0, where
>> a+b, a+c, a+d, b+c, b+d, c+d are all squares,
>> a+b+c, a+b+d, a+c+d, b+c+d are all cubes, and
>> a+b+c+d is a fourth power.
>>
>
> at least 3 of a,b,c,d would have to be divisible by 4
> at least 3 would have to be divisible by 5
> and
> at least 3 would have to be divisible by 7
>

Ignore my previous email - I was hand calculating an made several errors.
It is true that at least 3 of a,b,c,d would have to be divisible by 4,
but not so for  5 and 7.


the possible multisets congruent to {a,b,c,d} modulo 4 are
{0,0,0,0}
{0,0,0,1}

the possible multisets congruent to {a,b,c,d} modulo 5 are
{0,0,0,0}
{0,0,0,1}
{0,0,1,4}
{1,3,3,3}
{2,2,2,4}
{2,2,3,3}
{2,3,3,3}

the possible multisets congruent to {a,b,c,d} modulo 7 are
{0,0,0,0}
{0,0,0,1}
{2,2,2,2}

the possible multisets congruent to {a,b,c,d} modulo 9 are
{0,0,0,0}
{0,0,0,1}

Andrew