Number Theory. Argue Is not the square of an integer.

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SUMMARY

The expression (17^4)*(5^10)*(3^5) is not the square of an integer due to the presence of the prime factor 3 raised to an odd exponent. While 17^4 and 5^10 yield perfect squares (289 and 3125 respectively), the term 3^5 results in a non-integer when taking the square root, specifically sqrt(243) which equals approximately 15.5884. This demonstrates that for a product to be a perfect square, all prime factors must have even exponents.

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Homework Statement


Argue that (17^4)*(5^10)*(3^5) is not the square of an integer.



Homework Equations


N/A?


The Attempt at a Solution


Do I break these up, and show that each is not a square? I'm not sure if that would be correct, but sqrt(17^4)=289 * sqrt(5^10)=3125 * sqrt(243)=15.5884 =...

Since sqrt(243)=15.5884 and is not an integer then the above is not the square of an integer. Is this an efficient explanation?
 
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Well, it certainly doesn't show any understanding of the problem! Look at the exponents: [itex]\sqrt{17^4}= (17^4)^{1/2}= 17^2[/itex]. [itex]\sqrt{5^{10}}= (5^{10})^{1/2}= 5^5[/itex]. What about [itex]\sqrt{3^5}[/itex]?

Do you see why the fact that 3, 5, and 17 are prime numbers is important?
(Consider the same question about [itex]\sqrt{(8)(18)}[/itex].)
 

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