Ed,
"Find positive integers A, B, C, D, and E all less than 100 so that
A^2 + B^2 + C^2 = D^2 + E^2 and A^3 + B^3 + C^3 = D^3 + E^3"
Answer: For integers from 199, there is one unique solution set not
counting swapping A, B, C around, or swapping D and E: For A,B,C, the
numbers are 21, 26, 64. For D and E, the numbers are 37 and 62.
Counting all combinations of these due to addition reordering, there
are 12 solution combinations.
Further analysis: Counting all possible combinations, there are
647913 solutions to A^2 + B^2 + C^2 = D^2 + E^2, and 11550 solutions
to A^3 + B^3 + C^3 = D^3 + E^3.
If we change the range from 199 to 1199, there are three solution
sets (or 36 total solution combinations counting addition reordering):
1. (A,B,C)=42,52,128; (C,D)=74,124 2. (A,B,C)=63,78,192; (C,D)=
111,186 3. (A,B,C)=21,26,64; (C,D)=37,62
Thanks for the diversion on a slow work day!
Happy Holidays,
J.B. Gill


A, B, C, D, and E all less than 100 so that A2 + B2 + C2 = D2 + E2
and A3 + B3 + C3 = D3 + E3 are: A=21, B=26, C=64 or a permutation of
these. and D=37, E=62 or D=62 and E=37.
Regarding ending of The square of 40081787109376 which starts like
16065496578813... is in the same digit as the origional number i.e.
40081787109376 , so the complete square is
1606549657881340081787109376, it is a automorphic number.
Shyam Sunder Gupta www.shyamsundergupta.com guptass@rediffmail.com
 Hi Ed,
I was unable to find anything else than (21,26,64) and (37,62) as
solutions.
KR,
JeanFrançois Halleux


Erich Friedman: Find positive integers A, B, C, D, and E all less than
100 so that A2 + B2 + C2 = D2 + E2 and A3 + B3 + C3 = D3 + E3.
Solution: a=21 b=26 c=64 d=37 e=62
Dave Gao 
The only solution which I have found to the problem, with a hurried
program, is: 21^2 +26^2 +64^2 = 37^2 +62^2 = 5213 21^3 +26^3 +64^3 =
37^3 +62^3 = 288981
Happy New Year
Jordi Domènech
 My
answer to the puzzle A^2+B^2+C^2=D^2+E^2 and A^3+B^3+C^3=D^3+E^3 is as
follows:
A=21 B=26 C=64 D=37 E=62
I really enjoy your puzzle page!
Bill Estabrook

sorry, had a little floatingpointinaccuracy in my proggy..
21² + 26² + 64² = 37² + 62² 21³ + 26³ + 64³ = 37³ + 62³
Franz.
p.s.: can this be solved algebraic ?
 21^2 +
26^2 + 64^2 = 37^2 + 62^2 21^3 + 26^3 + 64^3 = 37^3 + 62^3 Who
invented that this is possible? Must be a genious!
Juha Saukkola

"Find positive integers A, B, C, D, and E all less than 100 so that
A^2 + B^2 + C^2 = D^2 + E^2 and A^3 + B^3 + C^3 = D^3 + E^3."
I think I have found the only answer.
A=21 B=26 C=64
D=37 E=62
Since A, B, & C are interchangeable, any of the values for those
variables can be switched (e.g. (A,B,C) = (26,21,64) is also
acceptable. The same is true for D & E.
I used Microsoft Excel to solve this one, and it took me hours.
Alan Lemm 
Find positive integers A, B, C, D, and E all less than 100 so that A2
+ B2 + C2 = D2 + E2 and A3 + B3 + C3 = D3 + E3.
Answer: A = 21, B = 26, C = 64, D = 37, E = 62
Daniel Lidström
 Solution:
21, 26, 64, 37, 62 are values for A, B, C, D, and E
that satisfy the two conditions:
A^2 + B^2 + C^2 = D^2 + E^2
and A^3 + B^3 + C^3 = D^3 + E^3
I used a brute force approach with Excel, first creating a million
item spread sheet containing all of the possible "D^3+E^3C^3," then
searching it for any results equivalent to one of the 10,000 possible
"A^3 + B^3" sums, and lastly testing the coordinates that yielded
matches in the squares equation.
I'm curious to know of more insightful mathier strategies, and what
patterns appear among the two sets of solutions to help find their
intersection. So back to the spreadsheets to see what I can see.
 Matt Sheppeck
While I was marvelling at the speed with which my Christmas laptop can
perform 100,000,000 comparisons, Tom Hanks was talking of amazing
roomsize NASA computers performing thousands of calculations in
Apollo 13.

200:3 > 400:0:9 > 1600:8:1 > 2:560065 > 3:1:36:7:2:8:0:4:22:9 >
2004 200:3 > 4000:9 > 1600:0:0:81 > 2:560065 >
3:1:36:7:2:8:0:4:22:9 > 2004 200:3 > 4000:9 > 16000:0:81 >
2:5:60065:6:1 > 3:6:0:7:8:0:42:9:1 > 2004 200:3 > 4000:9 >
1:600008:1 > 3600:0:9:600066 > 3:6:0:0:9:21:6:4:4:37 > 2004 200:3
> 4000:9 > 1600008:1 > 25:60:0:25:60006::5 > 36:0:0:7:24:9:1:1 >
2004 200:3 > 4000:9 > 1:6000081 > 3:6000:97:2006:5:6:2 >
40:0:3:3:5:19 > 2004 200:3 > 4000:9 > 160000:81 > 2:5:6000065:6:1
> 3:6:0:0:0:7:8:0:0:0:42:9:1 > 2004
And I'm sure there are a lot more =) Yogy Namara

Hi Ed
The Erich Friedman problem A^2 + B^2 + C^2 = D^2 + E^2 and A^3 + B^3 +
C^3=D^3 + E^3
Solution
A=21
B=26
C=64
D=37
E=62
Regards
Paul Cleary.