Suppose X
3Y
5Z
2 is a perfect seventh power, where each of X, Y, and Z
is a positive integer.
Is XY4Z3 always a perfect seventh power?
- If so, prove it.
- If not, provide a counterexample.
Computer proof:
ct=0;
for a=0:6
for b=0:6
for c=0:6
t=3*a+5*b+2*c;
t2=a+4*b+3*c;
if mod(t,7)==0
ct=ct+1;
disp([a b c t2])
if mod(t2,7)~=0
disp('***')
end
end
end
end
end
ct
Lists, for any given prime included in any of X, Y, or Z's prime factorization, each of the inclusions in those variables that allows the number to be a perfect seventh power.
For example, the lines
2 4 1 21 for prime 2
0 3 3 21 for prime 3
1 1 3 14 for prime 5
corresponds to X=2^2 * 5; Y= 2^4 * 3^3 * 5; Z = 2 * 3^3 * 5^3, where
X^3 * Y^5 * Z^2 = 2^28 * 3^21 * 5^42
so that the condition is met, and the result of the second formula is also a power of 7.
These are all mod 7. Therefore consider the last column below mod 7.
Powers of prime total power
in: in X*Y^4*Z^3
X Y Z
0 0 0 0
0 1 1 7
0 2 2 14
0 3 3 21
0 4 4 28
0 5 5 35
0 6 6 42
1 0 2 7
1 1 3 14
1 2 4 21
1 3 5 28
1 4 6 35
1 5 0 21
1 6 1 28
2 0 4 14
2 1 5 21
2 2 6 28
2 3 0 14
2 4 1 21
2 5 2 28
2 6 3 35
3 0 6 21
3 1 0 7
3 2 1 14
3 3 2 21
3 4 3 28
3 5 4 35
3 6 5 42
4 0 1 7
4 1 2 14
4 2 3 21
4 3 4 28
4 4 5 35
4 5 6 42
4 6 0 28
5 0 3 14
5 1 4 21
5 2 5 28
5 3 6 35
5 4 0 21
5 5 1 28
5 6 2 35
6 0 5 21
6 1 6 28
6 2 0 14
6 3 1 21
6 4 2 28
6 5 3 35
6 6 4 42
ct =
49
All 49 possibilities result in the second formula also being a power of 7, as the exponents add to a multiple of 7.
Formal summary:
For X^3*Y^5*Z^2 to be a perfect seventh power, each of its prime factors must fall into X, Y and Z according to one of the 49 patterns shown above. When they do, in each case X*Y^4*Z^3 also is a perfect seventh power.
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Posted by Charlie
on 2023-06-30 16:03:51 |
computer solution
