341x^2+7x+13 is a square \Rightarrow 4x^2+28x+52 = (2x+7)^2+3 is a perfect square.
If m2, n2 are both distinct greater than 4, their difference is greater than 4. Hence
we must have |2x+7|=1 which has solutions -3,-4
(1) find all Integral value of x for which the expression
x^{2}+7x+13
is a perfect square.
i have got only one solution x= - 4
is there is any other Integral value of x which satisfy this expression?
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5 Answers
341
1understand it this way
3=k^2-(2x+7)^2 \\ \Rightarrow 3=\left(k-(2x+7) \right)\left(k+\left(2x+7 \right) \right)\\ since \ 3 \ is \ prime \ \\ \begin {cases} k-\left(2x+7 \right)=3 \\ k+\left(2x+7 \right)=1\end{cases} \\ \begin {cases} k-\left(2x+7 \right)=1 \\ k+\left(2x+7 \right)=3\end{cases} \\ \begin {cases} k-\left(2x+7 \right)=-3 \\ k+\left(2x+7 \right)=-1\end{cases} \\ \begin {cases} k-\left(2x+7 \right)=-1 \\ k+\left(2x+7 \right)=-3\end{cases} \texttt{solve these equations to get x's}
341no that's not what I had in mind.
We are claiming that (2x+7)2+3 is a perfect square.
But (2x+7)2 is itself a perfect square.
If you look at the sequence of squares 0,1,4,9,16,25,... the difference between them even consecutive squares increasing.
Hence, the only squares whose difference is 3 are 1 and 4.
So |2x+7|=1