Does not $(100x+51)^2-1$ end in $4$ zeros?

Question

$$(100x+51)^2-1$$ For when $x$ is a real positive interger. Prove or disprove that the expression doea not end in four zeros or equivalently be in the equation $1000y$ (where $y$ is a real positive interger)

Answer 1

As has been pointed out in other answers,

$$(100x + 51)^2 - 1 = 10000x^2 + 10200x + 2600$$

So to find a value ending in 4 zeroes, you need $200x + 2600 = 10000$ , or $x = 37$ (plus any multiple of $50$).

Answer 2

You don't have to use modular arithmetic (if you are thinking about it).

$(100x+51)^2-1=10000x^2+2\times 100x\times 51+2601-1=10000x^2+10200x+2600$, definitely not necessary ending with $4$ zeroes.

With $x=37$: $$(100x+51)^2-1=(100\times37+51)^2-1=3751^2-1=14070001-1=14070000$$

Answer 3

Take $x=77$ then we have $$(100\cdot 77+51)^2-1=60078000$$ this ends with three Zeros.