Question

The equation x^2 = (y+1)^2 - y^2, where x and y are both positive integers has an infinite number of solutions.

(i) One solution is x=3, y=4. Find six other solutions.

(ii) Write down six solutions to the equation x^2 + y^2 = z^2, where x, y and z are all positive integers with no factors in common and where x 11 years ago

Answer 1

@zzr0ck3r

Answer 2

\(\large\tt \begin{align} \color{black}{\color{red}{1.)}\hspace{.33em}\\~\\
x^2=(y+1)^2-y^2\hspace{.33em}\\~\\
x^2=y^2+2y+1-y^2\hspace{.33em}\\~\\
x^2=2y+1\hspace{.33em}\\~\\
\dfrac{x^2-1}{2}=y\hspace{.33em}\\~\\
\dfrac{(x-1)(x+1)}{2}=y\hspace{.33em}\\~\\
\normalsize \text{here if u choose even integers,}\hspace{.33em}\\~\\
\normalsize \text{ there will be no integers solution for y}\hspace{.33em}\\~\\
\normalsize \text{so choose only odd integers above 3, }\hspace{.33em}\\~\\
\normalsize \text{as 3 is already given }\hspace{.33em}\\~\\
\normalsize \text{example }\hspace{.33em}\\~\\
\dfrac{(5-1)(5+1)}{2}=y\hspace{.33em}\\~\\
y=12
}\end{align}\)

Answer 3

looks neat xD that also proves that every odd integer can be expressed as difference of two squares

Answer 4

thnks

Answer 5

\(\large\tt \begin{align} \color{black}{ \color{red}{2.)}\hspace{.33em}\\~\\
\normalsize \text{for 2 }\hspace{.33em}\\~\\
\normalsize \text{we can use the same solution as 1 }\hspace{.33em}\\~\\
x^2 = (y+1)^2 - y^2 \hspace{.33em}\\~\\
x^2+y^2=(y+1)^2\hspace{.33em}\\~\\
\normalsize \text{put z=x+1 }\hspace{.33em}\\~\\
\normalsize \text{use the same process as stated in }~~\color{red}{1.)}\hspace{.33em}\\~\\
\normalsize \text{example,for x=5 }\hspace{.33em}\\~\\
\dfrac{(x-1)(x+1)}{2}=y\hspace{.33em}\\~\\
\dfrac{(5-1)(5+1)}{2}=y\hspace{.33em}\\~\\
y=12 \hspace{.33em}\\~\\
5^2+12^2=13^2\hspace{.33em}\\~\\
\gcd (5,12,13)=1\hspace{.33em}\\~\\
}\end{align}\)

Answer 6

Please note that, another way to answer to the question i) is this:
I call x with m, and y with n,
so I can write:
\[m ^{2}=(n+1)^{2}-n ^{2}=2n+1\]
So we have to find all perfect square and odd numbers.
For example, we have:
\[9 \; \rightarrow m=3,\quad n=4\]
\[25 \; \rightarrow m=5,\quad n=12\]
\[49 \; \rightarrow m=7,\quad n=24\]
\[81 \; \rightarrow m=9,\quad n=40\]
\[121 \; \rightarrow m=11,\quad n=60\]
\[169 \; \rightarrow m=13,\quad n=84\]
\[225 \; \rightarrow m=15,\quad n=112\]
Since there are infinite odd numbers, then there are infinite solutions.