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masha68 [24]
3 years ago
6

Prove that if an integer is a perfect square and a perfect cube, simultaneously, then it is either of the form 7k or 7k+1

Mathematics
1 answer:
Bogdan [553]3 years ago
5 0
If an integer is both a square and a cube, it can be of the form:
<span>(<span>a3</span><span>)^2</span></span>
Now,
since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath),
 we write
<span><span>a^3</span>=7k</span>
and get the no to be
49k^2
, which is in the form of 7 times something
<span>49<span>k^2</span>=7×(7<span>k^2</span>)</span>
Now put
<span><span>a^3</span>=7k+−1</span> Square it
and you'll get a number in the form of (7times something +1)
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