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my question

Find the numbers of the form $\overline{abcd}$ for which the relations are checked simultaneously:

i) $\overline{ab}$ and $\overline{cd}$ are consecutive natural numbers

ii) $(2*\overline{ab}+3)(2*\overline{ab}+7)=\overline{abcd}$.

my idea and the point I got stuck at

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The point I got stuck at is if $\overline{ab}=\overline{cd}+1$. Hope one of you can help me!!!

IONELA BUCIU
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1 Answers1

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To solve $4x^2-81x+22=0$, you could either use the quadratic formula, or since you only care about integer solutions, you can just try divisors of $22$. Neither $1,2,11,$ nor $22$ work, so it has no positive integer solutions.

Eric
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  • Alternative : $~(81)^2 - (79)^2 = 320.~$ So, you have that $~(81)^2 - [4 \times 4 \times 22] = (81)^2 - 352,~$ which is between $~(79)^2~$ and $~(78)^2.$ – user2661923 Oct 29 '23 at 12:18