Demonstrate that the digits of a perfect square cannot add to 2, 3, 5, 6, or 8

1 answer:

5 0

It depends on what did you mean by saying perfect square. If I've understood it correctly, I can help you with a part of your problem. The squares of mod 9 are 1,4,7 which are came from 1,2,4. Addition of the given numbers are 2,3,5,6, 8, which are exactly the part of your problem. This number, which is not shown as squares Mod 9, and thus doesn't appear as a sum of digits of a perfect square. I hope you will find it helpful.

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$h^2 = 39^2-34^2$
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Answer:

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Step-by-step explanation:

49x^[2] + 28x - 10 = p^[2] + 4p -10

This equation is in the form a^[2]x + bx + c.

The 'c' is common for both equations, this means the 'a' and 'b' must also be common.

There are two ways to find p: 'a' or 'b'

a method

49x^[2] = p^[2]

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b method

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Answer:

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