assume that in addition to the integers a,b element of Z there also exist c,d element of Z such that p=c^2+d^2. Prove that a^2=c^2 and b^2=d^2 or
Added by Daniel K.
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However, the statement as it stands does not provide enough context or conditions to directly conclude that \( a^2 = c^2 \) and \( b^2 = d^2 \). Let's assume that \( p \) is some integer that can be expressed as the sum of two squares, and we want to explore the Show more…
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Prove the statement: For all integers a, b,and c, if a2 + b2 = c2, then a or b is even.
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Therefore, if a^2 + b^2 = c^2, then a or b (or both!) must be even. Assume for a contradiction that a^2 + b^2 = c^2. but both a and b are odd. Assume for a contradiction that a^2 + b^2 = c^2. but a or b are odd. Then it follows that both a^2 and b^2 are odd, and so their sum, c^2, must be odd (and so c is odd). Consider now a = 2m + 1, b = 2n + 1, and c = 2k. It follows that a^2 + b^2 = c^2 ⟹ 4m^2 + 4n^2 + 4m + 4n + 2 = 4k^2 and so 2 = 4(k^2 - m^2 - n^2 - m - n). Assume that both a and b are odd so that we can show that a^2 + b^2 ≠c^2
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