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anonymous
 3 years ago
Is this true: 'The only way you can get from one perfect square, q^2, to another, p^2, by multiplying by a constant [i.e. p^2=a^2 q^2] is if q IS that factor [i.e. q^2=a^2]'?
anonymous
 3 years ago
Is this true: 'The only way you can get from one perfect square, q^2, to another, p^2, by multiplying by a constant [i.e. p^2=a^2 q^2] is if q IS that factor [i.e. q^2=a^2]'?

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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1'the first square IS that factor' what is 'that' here? factor of whose ? \(b^2=ma^2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[q^2=49p^2\]Must \[q=49\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I've edited it. Is that still ambiguous?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1then i don't think its always true.... q equal 49, is not necessary

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Of course apart from the trivial case where p=1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Counterexample: 6^2= 4 * 2^2 Damn, it's wrong!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This popped up in http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201210&t=mat&l=en if anyone's interested...
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