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space time
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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

#23205
Cheenta Support
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For $$n>3$$, 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.

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