Math Olympiad, I.S.I. Entrance and College Mathematics › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Number Theory › number theory and inequalities

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- September 8, 2018 at 2:38 pm #22510
a and b are natural numbers

(a^2 + b^2 -d)^n = ( a^n + b^n)^2

n > 2 , d is an whole number whose value ranges from 0 to ab

prove that no natural solutions exist

October 1, 2018 at 2:05 pm #23205For \(n>3\), if \(n\) is odd, then this means \(a^n+b^n\) is a perfect \(n^{th}\) power. By Fermat’s Last Theorem,

no positive integer solutions exist.If \(n\) is even, suppose that \(n=2k\). Then, let \(x=a^2\) and \(y=b^2\). The equation becomes:

\((x+y-d)^k=x^k+y^k\). By Fermat’s Last Theorem, no positive integer solutions exist as well.We now need to check \(k=1,2\). For the first, note that \(a^2+b^2<(a+b)^2\), so no solutions.

For the second,

\(a^2+b^2-d=a^2+b^2\) implies \(d=0\)

So any \(a,b\) work with \(d=0\). However, if the range of \(d\) is exclusive, there are no solutions.The solution is Provided by our teaching assistant Tarit Goswami.

- This reply was modified 4 months, 2 weeks ago by Cheenta Support.

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