+0

# Need Help!

0
479
2

What is the least positive integer value of $x$ such that $(2x)^2 + 2\cdot 37\cdot 2x + 37^2$ is a multiple of 47?

May 1, 2019

#1
+1

x=1; a=(((2 *x)^2 + 2*37*2* x + 37^2) % 47);printa," ",x; x++; if(x<=300, goto1, discard=0;

The smallest x = 5. Then every 47 numbers after that: 5, 52, 99, 146, 193, 240, 287.......and so on.

May 1, 2019
#2
+25460
+2

What is the least positive integer value of $$x$$ such that $$(2x)^2 + 2\cdot 37\cdot 2x + 37^2$$ is a multiple of $$47$$?

$$\begin{array}{|rcll|} \hline \mathbf{(2x)^2 + 2\cdot 37\cdot 2x + 37^2} &=& \mathbf{\left(2x+37\right)^2} \\\\ \left(2x+37\right)^2 &=& \left(47n\right)^2 \\ 2x+37 &=& 47n,\ \qquad n \in \mathbf{Z} \\ 2x &=& 47n - 37 \\\\ \mathbf{ x } &=& \mathbf{\dfrac{ 47n - 37 } {2}} \quad | \quad 47n - 37 \text{ is even, so } n \text{ is odd} \\\\ \mathbf{ x_{\text{least positive integer}} } &=& \mathbf{\dfrac{ 47n - 37 } {2}},\ \text{if } n = 1 \\\\ &=& \dfrac{ 47 - 37 } {2} \\\\ &=& \dfrac{ 10 } {2} \\\\ \mathbf{ x_{\text{least positive integer}} } &=& \mathbf{5} \\ \hline \end{array}$$

May 2, 2019