Question 178138
lets call our integers a,b=a+1,and c=a+2
:
{{{(a+2)^2=(a+1)^2+100}}}
:
{{{a^2+4a+4=a^2+2a+1+100}}} multiplied terms
:
{{{4a+4=2a+101}}}cancelled squared terms and combined like terms on each side
:
{{{2a=97}}}
:
{{{a=97/2}}}
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there are NO integers for which these stipulations are true
:
had the 2nd number been squared plus 101 instead of 100 then the integers would have been 49,50, and 51. I suspect someone just made a mistake when writing the problem or there was a mis-typed number
:
the rational number answers under these stipulations would be 
:
{{{system(97/2,99/2,101/2)}}}