the equation (m² + n²)² = 8r²s²
the problem: find integers for m, n, r & s
is this a fair enough way to prove the above is impossible
Code:
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(m² + n²)² = 8r²s²
(m² + n²)² = 2(4r²s²)
m² + n² = (2rs)(sqrt(2))
m² + n²
------- = sqrt(2)
2rs
since sqrt(2) is irrational, m, n, r & s can't all be integers
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No, because m = n = r = s = 0 works in the *given* equation.
Because each variable is raised to an even positive exponent,
then let us assume for relative simplicity that each of the
four variables represents strictly positive integers.**
Because it is difficult for me to see those relatively small exponents,
I will retype it:
\(\displaystyle (m^2 + n^2)^2 = 8r^2s^2\)
\(\displaystyle (m^2 + n^2)^2 = 2(4r^2s^2)\)
\(\displaystyle m^2 + n^2 = 2rs\sqrt{2}, \ \ because \ of \ **, \ don't \ use \ \pm.\)
\(\displaystyle \frac{m^2 + n^2}{2rs} = \sqrt{2}\)
Then the numerator and denominator are made up of integers,
based on an assumption. This fraction is a ratio of two
integers.
But because \(\displaystyle \sqrt{2}\) is known to be irrational,
then it cannot equal the ratio of two integers.
Then this is a contradiction, so the original statement is not true.
At least one of m, n , r, s must not be an integer.
(Potentially, none of them may be integers. However,
I am not claiming that that scenario is possible.)
