Find-the-number-of-triangles-with-integer-side-lengths-and-perimeter-p-

Question Number 188301 by mr W last updated on 28/Feb/23

$${Find}\:{the}\:{number}\:{of}\:{triangles}\:{with} \\ $$$${integer}\:{side}\:{lengths}\:{and}\:{perimeter}\:{p}. \\ $$

Commented by mr W last updated on 28/Feb/23

$${as}\:{Q}\mathrm{188239},\:{but}\:{in}\:{general}\:{case}. \\ $$

Commented by BaliramKumar last updated on 27/Feb/23

$${why} \\ $$

Commented by mr W last updated on 27/Feb/23

$${for}\:{more}\:{solutions}! \\ $$

Commented by mr W last updated on 27/Feb/23

$${if}\:{you}\:{have}\:{a}\:{solution},\:{please}\:{share}. \\ $$

Answered by mr W last updated on 04/Mar/23

this is my solution: say the sides of the triangle are a,b,c with a≤b≤c, then the question is to find the number of solutions of following equation a+b+c=p ...(i) with 1≤a≤b≤c and a+b>c. say a=1+u with u≥0 b=a+v=1+u+v with v≥0 c=b+w=1+u+v+w with w≥0 and (1+u)+(1+u+v)>(1+u+v+w) ⇒u>w−1 ⇒u≥w say u=w+s with s≥0 then the equation (i) becomes to (1+w+s)+(1+w+s+v)+(1+w+s+v+w)=p ⇒3+4w+3s+2v=p ...(I) with w,s,v≥0 the number of solutions of eqn. (I) is the coefficient of x^p term in the expansion of x^3 (1+x^4 +x^8 +...)(1+x^3 +x^6 +...)(1+x^2 +x^4 +...) i.e. (x^3 /((1−x^4 )(1−x^3 )(1−x^2 ))). its expansion see below. examples: p=16: the coef. of x^(16) is 5. ⇒ 5 triangles have p=16. p=50: the coef. of x^(50) is 52. ⇒ 52 triangles have p=50. p=75: the coef. of x^(75) is 127. ⇒ 127 triangles have p=75.

$${this}\:{is}\:{my}\:{solution}: \\ $$$${say}\:{the}\:{sides}\:{of}\:{the}\:{triangle}\:{are}\:{a},{b},{c} \\ $$$${with}\:{a}\leqslant{b}\leqslant{c},\:{then}\:{the}\:{question}\:{is}\:{to}\: \\ $$$${find}\:{the}\:{number}\:{of}\:{solutions}\:{of}\: \\ $$$${following}\:{equation} \\ $$$${a}+{b}+{c}={p}\:\:\:…\left({i}\right) \\ $$$${with}\:\mathrm{1}\leqslant{a}\leqslant{b}\leqslant{c}\:{and}\:{a}+{b}>{c}. \\ $$$${say} \\ $$$${a}=\mathrm{1}+{u}\:{with}\:{u}\geqslant\mathrm{0} \\ $$$${b}={a}+{v}=\mathrm{1}+{u}+{v}\:{with}\:{v}\geqslant\mathrm{0} \\ $$$${c}={b}+{w}=\mathrm{1}+{u}+{v}+{w}\:{with}\:{w}\geqslant\mathrm{0} \\ $$$${and} \\ $$$$\left(\mathrm{1}+{u}\right)+\left(\mathrm{1}+{u}+{v}\right)>\left(\mathrm{1}+{u}+{v}+{w}\right) \\ $$$$\Rightarrow{u}>{w}−\mathrm{1}\:\Rightarrow{u}\geqslant{w} \\ $$$${say}\:{u}={w}+{s}\:{with}\:{s}\geqslant\mathrm{0} \\ $$$${then}\:{the}\:{equation}\:\left({i}\right)\:{becomes}\:{to} \\ $$$$\left(\mathrm{1}+{w}+{s}\right)+\left(\mathrm{1}+{w}+{s}+{v}\right)+\left(\mathrm{1}+{w}+{s}+{v}+{w}\right)={p} \\ $$$$\Rightarrow\mathrm{3}+\mathrm{4}{w}+\mathrm{3}{s}+\mathrm{2}{v}={p}\:\:\:\:…\left({I}\right) \\ $$$${with}\:{w},{s},{v}\geqslant\mathrm{0} \\ $$$${the}\:{number}\:{of}\:{solutions}\:{of}\:{eqn}.\:\left({I}\right) \\ $$$${is}\:{the}\:{coefficient}\:{of}\:{x}^{{p}} \:{term}\:{in} \\ $$$${the}\:{expansion}\:{of} \\ $$$${x}^{\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{4}} +{x}^{\mathrm{8}} +…\right)\left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +…\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +…\right) \\ $$$${i}.{e}.\:\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}. \\ $$$${its}\:{expansion}\:{see}\:{below}. \\ $$$$ \\ $$$${examples}: \\ $$$${p}=\mathrm{16}:\:{the}\:{coef}.\:{of}\:{x}^{\mathrm{16}} \:{is}\:\mathrm{5}. \\ $$$$\Rightarrow\:\mathrm{5}\:{triangles}\:{have}\:{p}=\mathrm{16}. \\ $$$${p}=\mathrm{50}:\:\:{the}\:{coef}.\:{of}\:{x}^{\mathrm{50}} \:{is}\:\mathrm{52}. \\ $$$$\Rightarrow\:\mathrm{52}\:{triangles}\:{have}\:{p}=\mathrm{50}. \\ $$$${p}=\mathrm{75}:\:{the}\:{coef}.\:{of}\:{x}^{\mathrm{75}} \:{is}\:\mathrm{127}. \\ $$$$\Rightarrow\:\mathrm{127}\:{triangles}\:{have}\:{p}=\mathrm{75}. \\ $$

Commented by mr W last updated on 04/Mar/23