10000th User wrote: Let $a$ be an odd positive integer greater than 17 such that $3a-2$ is a perfect square. Show that there exist distinct positive integers $b$ and $c$ such that $a+b,a+c,b+c$ and $a+b+c$ are four perfect squares. $a=2k+1$, then $3a-2=6k+1=x^{2}$ Let $b=x^{2}-a=4k$ and $c=k^{2}-4k$. We have: $a+b=(2k+1)+(4k)=6k+1=x^{2}$ $b+c=(4k)+(k^{2}-4k)=k^{2}$ $a+c=(2k+1)+(k^{2}-4k)=k^{2}-2k+1=(k-1)^{2}$ $a+b+c=(2k+1)+(4k)+(k^{2}-4k)=k^{2}+2k+1=(k+1)^{2}$ we're done.

ElChapin wrote: 10000th User wrote: Let $a$ be an odd positive integer greater than 17 such that $3a-2$ is a perfect square. Show that there exist distinct positive integers $b$ and $c$ such that $a+b,a+c,b+c$ and $a+b+c$ are four perfect squares. $a=2k+1$, then $3a-2=6k+1=x^{2}$ Let $b=x^{2}-a=4k$ and $c=k^{2}-4k$. We have: $a+b=(2k+1)+(4k)=6k+1=x^{2}$ $b+c=(4k)+(k^{2}-4k)=k^{2}$ $a+c=(2k+1)+(k^{2}-4k)=k^{2}-2k+1=(k-1)^{2}$ $a+b+c=(2k+1)+(4k)+(k^{2}-4k)=k^{2}+2k+1=(k+1)^{2}$ we're done. But it says let $a$ be an odd positive integer greater than 17, that's got to mean something in the problem, the way I was doing it goes like this. $a=2n+1$ then $n > 8$ $3a-2=6n+1=k^{2}$. Therefore I have. $6n+1 \equiv 0\pmod{4}$ or $6n+1 \equiv 1\pmod{4}$ . Then $6n \equiv 0\pmod{4}$, the other case is not valid since $6n$ cannot left residuum 3 when dividing by 4. Solving that I got $n$ is even, so let's say $n=2d$ and $d > 4$. We have $a=4d + 1$ and $k^2=12d + 1$ and then I basically did the same as you did. $b=4d^2 - 8d$ and $c=8d$. I don't see what the statement $a > 17$ has to do with the problem solution. If anyone sees it please let me know...

10000th User wrote: Let $a$ be an odd positive integer greater than 17 such that $3a-2$ is a perfect square. Show that there exist distinct positive integers $b$ and $c$ such that $a+b,a+c,b+c$ and $a+b+c$ are four perfect squares. $b=2a-2$ and $c=\frac{a^2-10a+9}{4}$ works.

panamath wrote: ElChapin wrote: 10000th User wrote: Let $a$ be an odd positive integer greater than 17 such that $3a-2$ is a perfect square. Show that there exist distinct positive integers $b$ and $c$ such that $a+b,a+c,b+c$ and $a+b+c$ are four perfect squares. $a=2k+1$, then $3a-2=6k+1=x^{2}$ Let $b=x^{2}-a=4k$ and $c=k^{2}-4k$. We have: $a+b=(2k+1)+(4k)=6k+1=x^{2}$ $b+c=(4k)+(k^{2}-4k)=k^{2}$ $a+c=(2k+1)+(k^{2}-4k)=k^{2}-2k+1=(k-1)^{2}$ $a+b+c=(2k+1)+(4k)+(k^{2}-4k)=k^{2}+2k+1=(k+1)^{2}$ we're done. But it says let $a$ be an odd positive integer greater than 17, that's got to mean something in the problem, the way I was doing it goes like this. $a=2n+1$ then $n > 8$ $3a-2=6n+1=k^{2}$. Therefore I have. $6n+1 \equiv 0\pmod{4}$ or $6n+1 \equiv 1\pmod{4}$ . Then $6n \equiv 0\pmod{4}$, the other case is not valid since $6n$ cannot left residuum 3 when dividing by 4. Solving that I got $n$ is even, so let's say $n=2d$ and $d > 4$. We have $a=4d + 1$ and $k^2=12d + 1$ and then I basically did the same as you did. $b=4d^2 - 8d$ and $c=8d$. I don't see what the statement $a > 17$ has to do with the problem solution. If anyone sees it please let me know... If a=17 then d=4, which would imply b=c, but the problem requires b and c to be distinct. That's why a>17.

ElChapin wrote: 10000th User wrote: Let $a$ be an odd positive integer greater than 17 such that $3a-2$ is a perfect square. Show that there exist distinct positive integers $b$ and $c$ such that $a+b,a+c,b+c$ and $a+b+c$ are four perfect squares. $a=2k+1$, then $3a-2=6k+1=x^{2}$ Let $b=x^{2}-a=4k$ and $c=k^{2}-4k$. We have: $a+b=(2k+1)+(4k)=6k+1=x^{2}$ $b+c=(4k)+(k^{2}-4k)=k^{2}$ $a+c=(2k+1)+(k^{2}-4k)=k^{2}-2k+1=(k-1)^{2}$ $a+b+c=(2k+1)+(4k)+(k^{2}-4k)=k^{2}+2k+1=(k+1)^{2}$ we're done. How we can assume that $b+c=k^{2}$ if $b=4k$?