Question 934943
let sides be {{{a}}}, {{{b}}} and {{{c}}}

If it is a right angle triangle, then the relationship between the sides must follow the Pythagorean theorem:

{{{a^2 + b^2 = c^2}}}

Obviously, the side called {{{c}}} must be the longest one (because squares can never be negative).

Let's get rid of the "degenerate" solution of

{{{a = 7}}}
{{{b = 0}}}
{{{c = 7}}}

(this is called a "degenerate" triangle because one side measures {{{0}}})
(mathematically, it could still be a triangle, but in real life, it is simply a line segment)

You are looking for a solution where one side is {{{7}}} and the other two sides are {{{whole}}} {{{numbers}}}.

We know that there are no solutions if you use {{{c = 7}}}


Next, you can check values where {{{a = 7}}}
(you don't need to check {{{b=7}}} because it will be the same exercise)

{{{7^2 + b^2 = c^2}}}
{{{49 = c^2 - b^2}}}

The "formula" to find the difference between two squared integers goes like this

{{{c^2 - b^2 = c + b + (twice_ each_ number_ between_ b_ and_ c)}}}

for example

{{{17^2 - 16^2 = 17 + 16 = 33}}}

let's check

{{{289 - 256 = 33}}}

{{{20^2 - 17^2 = 20 + 17 + 18+18 + 19+19 = 111}}}

{{{400 - 289 = 111}}}

(this is not a coincidence, there is a true reason why this "trick" works).

So we need two numbers such that the difference of their squares is {{{49}}}

{{{c^2 - b^2 = 49}}}

because we need the two end numbers, the smallest difference would be {{{24}}} and {{{25}}}

because {{{24 + 25 = 49}}}

{{{25^2 - 24^2 = 625 - 576 = 49}}}


let's try

{{{7^2 + 24^2 = 25^2}}}

{{{49+ 576 = 625}}}

{{{625 = 625}}}

There is a triangle {{{7}}}, {{{24}}}, {{{25}}} would be a right-angle triangle with three sides being {{{whole}}}{{{ numbers}}} and one side being {{{7}}}.