Maths exam question on consecutive Pythagoras triplets Our 13 year old daughter brought home this maths question that she was asked in a "maths challenge" last week:
Q. The values of the adjacent and opposite sides in a right angle triangle of lengths 3, 4, 5 are consecutive. Obtain another triangle with consecutive adjacent and opposite sides having values between 500 and 1000.
My wife and I spent two hours trying to figure it out before we gave up. The teacher provided the solution (696, 697, 985) but could not explain how to obtain it. Searching on internet we can see solutions to this but none of them that we would expect even a very clever 13 year old to know. Are we missing something obvious?
Thanks,
Adrian

$x^2+(x+1)^2=y^2 \implies (x+1)^2 = (y+x)(y-x)$
Note that $a=y+x$ and $b=y-x$ gives $x={a-b\over 2}$ and the equation becomes $(a-b+2)^2=4ab$
$2x^2 > 500^2$, $2(x+1)^2 < 1000^2\implies x \geq 354$ and $x \leq 706$ so $1706\geq a\geq 854$ and $646 \geq b\geq 0$
Let $gcd(a,b)=k$ then $(a_1k-b_1k+2)^2=4a_1b_1k^2$ meaning $k^2 \mid (a_1k-b_1k+2)^2$ meaning $k \mid a_1k-b_1k+2$ meaning $k \mid 2$.
($1$) $k=1$ then $(a-b+2)^2=4ab$ where $a,b$ are coprime and are both square numbers.
 $42^2>1706\geq a\geq 854>29^2$ and $26^2>646 \geq b\geq 0$ means there are only $12$ values of $a$ to be checked. Furthermore $a$ cannot be even because if so then $b$ is odd to be coprime with $a$ and the left side becomes odd. So we  only needs to check $6$ values of $a$. 
Expand equation $b^2-2(3a+2)b+(a+2)^2=0$, $\Delta = 32a(a+1)$ is a square number or simply ${a+1\over 2}$ is a square number.
Plug in values $a=31^2,33^2,...,41^2$ and check ${a+1\over 2}$ we get only one pair ($a=41^2, b=17^2$), this correspond to the poster's $x=696$ 
($2$) $k=2$ then $(a_1-b_1+1)^2=16a_1b_1$, thus $a_1, b_1$ are both square numbers. $20^2<427\leq a_1\leq853<30^2, $$b_1 \leq 323<18^2$
Expand equation ${b_1}^2-2(9a_1+1)b_1+(a_1+1)^2=0$, $\Delta = 64a_1(5a_1+1)$ is a square number or simply $5a+1$ is a square number.
This time we have $9$ values of $a_1$ to check from $21$ to $29$. However none of the values produce an integral $b_1$. As a result $x=696$, given by the poster, is the only solution.
