/prog/ - arithmetic progression in c

Name: Anonymous 2011-01-05 1:13

Hi guys do have any ideas on how to make this arithmetic progression in c?
n^2 - m^2 - k^2 = aaa (a is any number)
for example 53^2 - 37^2 - 21^2 = 999
I need to find all the possible aaa's here

Name: Anonymous 2011-01-05 1:46

int n,m,k,a;for(n=0;n<256;n++)for(m=0;m<256;m++)for(k=0;k<256;k++)for(a=0;a<256;a++)assert((pow(n,2)-pow(m,2)-pow(k,2))!=a);

Name: Anonymous 2011-01-05 2:21

>>2
I think his specifications meant that it could be "a = 1" thus that "aaa = 111" or "a = 11" thus that "aaa = 111111."

Also, that looks painful.

Name: Anonymous 2011-01-05 2:33

>>3

long findaaa(long a){

int d=10;

while(1){

  if(a>=d){d*=10;break;} else return a*d+a*d*d+a*d*d*d;}}



long n,m,k,a,aaa=0;

for(n=1;n<256;n++)

 for(m=1;m<256;m++)

  for(k=1;k<256;k++)

   for(a=1,aaa;a<256;a++)

   { aaa=findaaa(a); assert((pow(n,2)-pow(m,2)-pow(k,2))!=aaa); }

Name: Anonymous 2011-01-05 2:35

>>4
better:



long findaaa(long a){

int d=10;

while(1){

  if(a>=d)d*=10;else return a*d+a*d*d+a*d*d*d;}}



long n,m,k,a;

for(n=1;n<256;n++)

for(m=1;m<256;m++)

for(k=1;k<256;k++)

for(a=1;a<256;a++)

assert((pow(n,2)-pow(m,2)-pow(k,2))!=(findaaa(a)));

Name: Anonymous 2011-01-05 2:40

>>5



long findaaa(long a){

int d=10;

while(1){

if(a>=d)d*=10;else return a+a*d+a*d*d;}}



long n,m,k,a;

for(n=1;n<256;n++)

for(m=1;m<256;m++)

for(k=1;k<256;k++)

for(a=1;a<256;a++)

assert((pow(n,2)-pow(m,2)-pow(k,2))!=(findaaa(a)));

Name: Anonymous 2011-01-05 6:55

progression in c
FOUR MEASURES Cm D#m Gm F THREAD OVER

Name: Anonymous 2011-01-05 8:01

>>7
I laughed heartily.

Name: Anonymous 2011-01-05 16:12

>>7
I hardly laughed

arithmetic progression in c

1 Name: Anonymous 2011-01-05 1:13

2 Name: Anonymous 2011-01-05 1:46

3 Name: Anonymous 2011-01-05 2:21

4 Name: Anonymous 2011-01-05 2:33

5 Name: Anonymous 2011-01-05 2:35

6 Name: Anonymous 2011-01-05 2:40

7 Name: Anonymous 2011-01-05 6:55

8 Name: Anonymous 2011-01-05 8:01

9 Name: Anonymous 2011-01-05 16:12