Question 694702
I do not see the whole elegant solution, but maybe this helps.
A sum of consecutive house numbers is the sum of an arithmetic sequence/progression, which can be calculated as the average of the first and last terms, times the number of terms.
The sum of the numbers of the houses to one side is
1+2+3+ ... +(x-2)+(x-1)={{{(1+(x-1))(x-1)/2=x(x-1)/2}}}
The sum of the numbers of the houses to the other side is
(x+1)+(x+2)+ ... +(n-1)+n={{{(x+1+n)(n-x)/2}}}
 
So {{{x(x-1)/2=(x+1+n)(n-x)/2}}} --> {{{x(x-1)=(x+1+n)(n-x)}}} --> {{{x^2-x=n^2+n-n^2-x}}} --> {{{2x^2=n^2+n}}} --> {{{2x^2=n(n+1)}}}
Since they are consecutive integers,
{{{n}}} and {{{n+1}}} cannot have any common factors.
For their product to be twice a perfect square, we need two perfect squares with no common factors:
one perfect square will be a factor of {{{n}}},
and another perfect square will be a factor of {{{n+1}}}
Calling the larger one {{{a^2}}},
either {{{system(n=a^2,(n+1)/2=b^2)}}} --> {{{2b^2=a^2+1}}} <--> {{{a^2=2b^2-1}}},
or {{{system(n/2=b^2,n+1=a^2)}}} --> {{{2b^2+1=a^2}}},
for some mutually prime positive integers {{{a}}} and {{{b}}}, with {{{a>b}}}
whose squares must be between {{{25=50/2}}} and {{{501}}}
Since {{{sqrt(25)=5}}} and {{{sqrt(501)<23}}}, we should look at numbers from 5 to 22.
As the larger square {{{a^2}}} is an odd number {{{a^2=2b^2 +- 1}}}, {{{a}}} is an odd number, and cannot be 22, so we should look at numbers from 5 to 22.
We are looking for a pair such that the square of the larger number is one more or one less than double the square of the smaller number.
We try numbers from 5 on as the smaller number {{{b}}}, and see what works.
I tried them making a table, which is hard for me to render here, so I will just show the calculations.
{{{b=5}}} --> {{{2b^2=2*5^2=50}}} --> {{{2b^2-1=49<50}}} and {{{2b^2+1=51}}}=not a square
{{{b=6}}} --> {{{2b^2=2*6^2=72}}} --> {{{2b^2-1=71}}}=not a square and {{{2b^2+1=73}}}=not a square
{{{b=7}}} --> {{{2b^2=2*7^2=98}}} --> {{{2b^2-1=97}}}=not a square and {{{2b^2+1=99}}}=not a square
{{{b=8}}} --> {{{2b^2=2*8^2=128}}} --> {{{2b^2-1=127}}}=not a square and {{{2b^2+1=129}}}=not a square
{{{b=9}}} --> {{{2b^2=2*9^2=162}}} --> {{{2b^2-1=161}}}=not a square and {{{2b^2+1=163}}}=not a square
{{{b=10}}} --> {{{2b^2=2*10^2=200}}} --> {{{2b^2-1=199}}}=not a square and {{{2b^2+1=201}}}=not a square
{{{b=11}}} --> {{{2b^2=2*11^2=242}}} --> {{{2b^2-1=241}}}=not a square and {{{2b^2+1=243}}}=not a square
{{{b=12}}} --> {{{2b^2=2*6^2=288}}} --> {{{2b^2-1=287}}}=not a square and {{{2b^2+1=highlight(289=17^2)}}}
The pair {{{b=12}}} and {{{a=17}}} with {{{a^2=2b^2+1}}} , works with {{{system(n/2=b^2,n+1=a^2)}}},
so {{{n+1=a^2}}} --> {{{n+1=17^2}}} --> {{{n+1=289}}} --> {{{highlight(n=288)}}}
and {{{2x^2=n(n+1)}}} --> {{{2x^2=288*289}}} --> {{{x^2=288*289/2}}} --> {{{x^2=
41616}}} --> {{{highlight(x=204)}}}
 
Since we expect only one solution, we could stop here.
Otherwise we would try further to see that nothing else works.
{{{b=13}}} --> {{{2b^2=2*13^2=339}}} --> {{{2b^2-1=337}}}=not a square and {{{2b^2+1=339}}}=not a square
{{{b=14}}} --> {{{2b^2=2*14^2=392}}} --> {{{2b^2-1=391}}}=not a square and {{{2b^2+1=393}}}=not a square
{{{b=15}}} --> {{{2b^2=2*15^2=450}}} --> {{{2b^2-1=449}}}=not a square and {{{2b^2+1=451}}}=not a square
{{{b>=16}}} --> {{{2b^2>=2*16^2=512}}} --> {{{2b^2+1>2b^2-1>=511>501}}}=not a number that could be {{{n}}} or {{{n+1}}}