Two parts of one question. Prove that when a perfect square number is divided by 7, the remainder cannot be possibly be 3, 5, 6. Therefore, is it possible using 2, 3, 4, 5, 6 to form a 5 digit number without repeating digits, and that the number is a perfect square?

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Anyinteger number can be expressed as

`7x +a`   where `a` is integer `in (0,6)`

Since all perfect squares are the square of particular integer numbers then all perfect squares `P` can be expressed as

`P = (7x+a)^2`

Now, expanding this out we get

`P = 49x^2 + 14ax + a^2 = 7(7x^2 + 2ax) + a^2`

The remainder upon dividing `P` by 7 will then be `a^2` modulo 7.

Since ` a` can only take the values 0,1,2,3,4,5,6 then `a^2` can only take the values 0,1,4,9,16,25,36 and `a^2` modulo 7 can only take the values
 0,1,4,2,2,4,1.

Therefore the remainder upon dividing a perfect square `P` by 7 can only take the values 0,1,2,4 unlike other integers that can also take the values 3,5,6.

Answer as proved above.

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